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justus27

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numina-1.5-numeric

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Problem 4. A $5 \times 100$ table is divided into 500 unit square cells, where $n$ of them are coloured black and the rest are coloured white. Two unit square cells are called adjacent if they share a common side. Each of the unit square cells has at most two adjacent black unit square cells. Find the largest possible value of $n$.

Solution. If we colour all the cells along all edges of the board together with the entire middle row except the second and the last-but-one cell, the condition is satisfied and there are 302 black cells. The figure below exhibits this colouring for the $5 \times 8$ case.  We can cover the table by one fragment like the first one on the figure below, 24 fragments like the middle one, and one fragment like the third one.  In each fragment, among the cells with the same letter, there are at most two coloured black, so the total number of coloured cells is at most $(5+24 \cdot 6+1) \cdot 2+2=302$.

302

Combinatorics

math-word-problem

Yes

Yes

olympiads

Problem 2. Let $n$ three-digit numbers satisfy the following properties: (1) No number contains the digit 0 . (2) The sum of the digits of each number is 9 . (3) The units digits of any two numbers are different. (4) The tens digits of any two numbers are different. (5) The hundreds digits of any two numbers are different. Find the largest possible value of $n$.

Solution. Let $S$ denote the set of three-digit numbers that have digit sum equal to 9 and no digit equal to 0 . We will first find the cardinality of $S$. We start from the number 111 and each element of $S$ can be obtained from 111 by a string of 6 A's (which means that we add 1 to the current digit) and $2 G$ 's (which means go to the next digit). Then for example 324 can be obtained from 111 by the string AAGAGAAA. There are in total $$ \frac{8!}{6!\cdot 2!}=28 $$ such words, so $S$ contains 28 numbers. Now, from the conditions (3), (4), (5), if $\overline{a b c}$ is in $T$ then each of the other numbers of the form $\overline{* c}$ cannot be in $T$, neither $\overline{* *}$ can be, nor $\overline{a * *}$. Since there are $a+b-2$ numbers of the first category, $a+c-2$ from the second and $b+c-2$ from the third one. In these three categories there are $$ (a+b-2)+(b+c-2)+(c+a-2)=2(a+b+c)-6=2 \cdot 9-6=12 $$ distinct numbers that cannot be in $T$ if $\overline{a b c}$ is in $T$. So, if $T$ has $n$ numbers, then $12 n$ are the forbidden ones that are in $S$, but each number from $S$ can be a forbidden number no more than three times, once for each of its digits, so $$ n+\frac{12 n}{3} \leq 28 \Longleftrightarrow n \leq \frac{28}{5} $$ and since $n$ is an integer, we get $n \leq 5$. A possible example for $n=5$ is $$ T=\{144,252,315,423,531\} $$ Comment by PSC. It is classical to compute the cardinality of $S$ and this can be done in many ways. In general, the number of solutions of the equation $$ x_{1}+x_{2}+\cdots+x_{k}=n $$ in positive integers, where the order of $x_{i}$ matters, is well known that equals to $\binom{n-1}{k-1}$. In our case, we want to count the number of positive solutions to $a+b+c=9$. By the above, this equals to $\binom{9-1}{3-1}=28$. Using the general result above, we can also find that there are $a+b-2$ numbers of the form $\overline{* c}$.

5

Combinatorics

math-word-problem

Yes

Yes

olympiads

## Problem 2 Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$, that touches $k_{1}$ and $k_{2}$ at $M$ and $N$, respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.

## Solution 1 Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 1). Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $A P$ as its base, so $\angle N A P=\angle N P A$. We have $\angle B A P=\angle B A M=\angle B M N$ and $\angle B A N=\angle B N M$. Thus we have $$ 180^{\circ}-\angle N B M=\angle B N M+\angle B M N=\angle B A N+\angle B A P=\angle N A P=\angle N P A $$ so the quadrangle $M B N P$ is cyclic (since the points $B$ and $P$ lie on different sides of $M N$ ). Hence $\angle A P B=\angle M P B=\angle M N B$ and the triangles $A P B$ and $M N B$ are congruent ( $M N=2 A M=A M+M P=A P$ ). From that we get $A B=M B$, i.e. the triangle $A M B$ is isosceles, and since $t$ is tangent to $k_{1}$ and perpendicular to $A M$, the centre of $k_{1}$ is on $A M$, hence $A M B$ is a right-angled triangle. From the last two statements we infer $\angle A M B=45^{\circ}$, and so $\angle N M B=90^{\circ}-\angle A M B=45^{\circ}$.  Figure 1

45

Geometry

math-word-problem

Yes

Yes

olympiads

Problem 2. For any set $A=\left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}$ of five distinct positive integers denote by $S_{A}$ the sum of its elements, and denote by $T_{A}$ the number of triples $(i, j, k)$ with $1 \leqslant i<j<k \leqslant 5$ for which $x_{i}+x_{j}+x_{k}$ divides $S_{A}$. Find the largest possible value of $T_{A}$.

Solution. We will prove that the maximum value that $T_{A}$ can attain is 4 . Let $A=$ $\left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}$ be a set of five positive integers such that $x_{1}x_{4}$ and $x_{3}>x_{2}$. Analogously we can show that any triple of form $(x, y, 5)$ where $y>2$ isn't good. By above, the number of good triples can be at most 5 and only triples $(1,2,5),(2,3,4)$, $(1,3,4),(1,2,4),(1,2,3)$ can be good. But if triples $(1,2,5)$ and $(2,3,4)$ are simultaneously good we have that: $$ x_{1}+x_{2}+x_{5} \mid x_{3}+x_{4} \Rightarrow x_{5}<x_{3}+x_{4} $$ and $$ x_{2}+x_{3}+x_{4} \mid x_{1}+x_{5} \Rightarrow x_{2}+x_{3}+x_{4} \leqslant x_{1}+x_{5} \stackrel{(1)}{<} x_{1}+x_{3}+x_{4}<x_{2}+x_{3}+x_{4}, $$ which is impossible. Therefore, $T_{A} \leqslant 4$. Alternatively, one can prove the statement above by adding up the two inequalities $x_{1}+x_{2}+x_{4}<x_{3}+x_{4}$ and $x_{2}+x_{3}+x_{4}<x_{1}+x_{5}$ that are derived from the divisibilities. To show that $T_{A}=4$ is possible, consider the numbers $1,2,3,4,494$. This works because $6|498,7| 497,8 \mid 496$, and $9 \mid 495$. Remark. The motivation for construction is to realize that if we choose $x_{1}, x_{2}, x_{3}, x_{4}$ we can get all the conditions $x_{5}$ must satisfy. Let $S=x_{1}+x_{2}+x_{3}+x_{4}$. Now we have to choose $x_{5}$ such that $$ S-x_{i} \mid x_{i}+x_{5} \text {, i.e. } x_{5} \equiv-x_{i} \quad \bmod \left(S-x_{i}\right) \forall i \in\{1,2,3,4\} $$ By the Chinese Remainder Theorem it is obvious that if $S-x_{1}, S-x_{2}, S-x_{3}, S-x_{4}$ are pairwise coprime, such $x_{5}$ must exist. To make all these numbers pairwise coprime it's natural to take $x_{1}, x_{2}, x_{3}, x_{4}$ to be all odd and then solve mod 3 issues. Fortunately it can be seen that $1,5,7,11$ easily works because $13,17,19,23$ are pairwise coprime. However, even without the knowledge of this theorem it makes sense intuitively that this system must have a solution for some $x_{1}, x_{2}, x_{3}, x_{4}$. By taking $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=$ $(1,2,3,4)$ we get pretty simple system which can be solved by hand rather easily.

4

Combinatorics

math-word-problem

Yes

Yes

olympiads

Problem 4. Let $M$ be a subset of the set of 2021 integers $\{1,2,3, \ldots, 2021\}$ such that for any three elements (not necessarily distinct) $a, b, c$ of $M$ we have $|a+b-c|>10$. Determine the largest possible number of elements of $M$.

Solution. The set $M=\{1016,1017, \ldots, 2021\}$ has 1006 elements and satisfies the required property, since $a, b, c \in M$ implies that $a+b-c \geqslant 1016+1016-2021=11$. We will show that this is optimal. Suppose $M$ satisfies the condition in the problem. Let $k$ be the minimal element of $M$. Then $k=|k+k-k|>10 \Rightarrow k \geqslant 11$. Note also that for every $m$, the integers $m, m+k-10$ cannot both belong to $M$, since $k+m-(m+k-10)=10$. Claim 1: $M$ contains at most $k-10$ out of any $2 k-20$ consecutive integers. Proof: We can partition the set $\{m, m+1, \ldots, m+2 k-21\}$ into $k-10$ pairs as follows: $$ \{m, m+k-10\},\{m+1, m+k-9\}, \ldots,\{m+k-11, m+2 k-21\} $$ It remains to note that $M$ can contain at most one element of each pair. Claim 2: $M$ contains at most $[(t+k-10) / 2]$ out of any $t$ consecutive integers. Proof: Write $t=q(2 k-20)+r$ with $r \in\{0,1,2 \ldots, 2 k-21\}$. From the set of the first $q(2 k-20)$ integers, by Claim 1 at most $q(k-10)$ can belong to $M$. Also by claim 1, it follows that from the last $r$ integers, at $\operatorname{most} \min \{r, k-10\}$ can belong to $M$. Thus, - If $r \leqslant k-10$, then at most $$ q(k-10)+r=\frac{t+r}{2} \leqslant \frac{t+k-10}{2} \text { integers belong to } M $$ - If $r>k-10$, then at most $$ q(k-10)+k-10=\frac{t-r+2(k-10)}{2} \leqslant \frac{t+k-10}{2} \text { integers belong to } M $$ By Claim 2, the number of elements of $M$ amongst $k+1, k+2, \ldots, 2021$ is at most $$ \left[\frac{(2021-k)+(k-10)}{2}\right]=1005 $$ Since amongst $\{1,2, \ldots, k\}$ only $k$ belongs to $M$, we conclude that $M$ has at most 1006 elements as claimed.

1006

Combinatorics

math-word-problem

Yes

Yes

olympiads

Problem 2. Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of the expression $$ A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c} $$  $19^{\text {th }}$ Junior Balkan Mathematical Olympiad June 24-29, 2015, Belgrade, Serbia

## Solution: We can rewrite $A$ as follows: $$ \begin{aligned} & A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-a^{2}-b^{2}-c^{2}= \\ & 2\left(\frac{a b+b c+c a}{a b c}\right)-\left(a^{2}+b^{2}+c^{2}\right)=2\left(\frac{a b+b c+c a}{a b c}\right)-\left((a+b+c)^{2}-2(a b+b c+c a)\right)= \\ & 2\left(\frac{a b+b c+c a}{a b c}\right)-(9-2(a b+b c+c a))=2\left(\frac{a b+b c+c a}{a b c}\right)+2(a b+b c+c a)-9= \\ & 2(a b+b c+c a)\left(\frac{1}{a b c}+1\right)-9 \end{aligned} $$ Recall now the well-known inequality $(x+y+z)^{2} \geq 3(x y+y z+z x)$ and set $x=a b, y=b c, z=c a$, to obtain $(a b+b c+c a)^{2} \geq 3 a b c(a+b+c)=9 a b c$, where we have used $a+b+c=3$. By taking the square roots on both sides of the last one we obtain: $$ a b+b c+c a \geq 3 \sqrt{a b c} $$ Also by using AM-GM inequality we get that $$ \frac{1}{a b c}+1 \geq 2 \sqrt{\frac{1}{a b c}} $$ Multiplication of (1) and (2) gives: $$ (a b+b c+c a)\left(\frac{1}{a b c}+1\right) \geq 3 \sqrt{a b c} \cdot 2 \sqrt{\frac{1}{a b c}}=6 $$ So $A \geq 2 \cdot 6-9=3$ and the equality holds if and only if $a=b=c=1$, so the minimum value is 3.

3

Inequalities

math-word-problem

Yes

Yes

olympiads

## Problem 4. An $L$-figure is one of the following four pieces, each consisting of three unit squares:  A $5 \times 5$ board, consisting of 25 unit squares, a positive integer $k \leq 25$ and an unlimited supply $L$-figures are given. Two players, $\boldsymbol{A}$ and $\boldsymbol{B}$, play the following game: starting with $\boldsymbol{A}$ they alternatively mark a previously unmarked unit square until they marked a total of $k$ unit squares. We say that a placement of $L$-figures on unmarked unit squares is called good if the $L$-figure do not overlap and each of them covers exactly three unmarked unit squares of the board. $\boldsymbol{B}$ wins if every $\boldsymbol{g o o d}$ placement of $L$-figures leaves uncovered at least three unmarked unit squares. Determine the minimum value of $k$ for which $\boldsymbol{B}$ has a winning strategy.

## Solution: We will show that player $\boldsymbol{A}$ wins if $k=1,2,3$, but player $\boldsymbol{B}$ wins if $k=4$. Thus the smallest $k$ for which $\boldsymbol{B}$ has a winning strategy exists and is equal to 4 . If $k=1$, player $\boldsymbol{A}$ marks the upper left corner of the square and then fills it as follows.  ## $19^{\text {th }}$ Junior Balkan Mathematical Olympiad June 24-29, 2015, Belgrade, Serbia If $k=2$, player $\boldsymbol{A}$ marks the upper left corner of the square. Whatever square player $\boldsymbol{B}$ marks, then player $\boldsymbol{A}$ can fill in the square in exactly the same pattern as above except that he doesn't put the $L$-figure which covers the marked square of $\boldsymbol{B}$. Player $\boldsymbol{A}$ wins because he has left only two unmarked squares uncovered. For $k=3$, player $\boldsymbol{A}$ wins by following the same strategy. When he has to mark a square for the second time, he marks any yet unmarked square of the $L$-figure that covers the marked square of $\boldsymbol{B}$. Let us now show that for $k=4$ player $\boldsymbol{B}$ has a winning strategy. Since there will be 21 unmarked squares, player $\boldsymbol{A}$ will need to cover all of them with seven $L$-figures. We can assume that in his first move, player $\boldsymbol{A}$ does not mark any square in the bottom two rows of the chessboard (otherwise just rotate the chessboard). In his first move player $\boldsymbol{B}$ marks the square labeled 1 in the following figure.  If player $\boldsymbol{A}$ in his next move does not mark any of the squares labeled 2,3 and 4 then player $\boldsymbol{B}$ marks the square labeled 3 . Player $\boldsymbol{B}$ wins as the square labeled 2 is left unmarked but cannot be covered with an $L$-figure. If player $\boldsymbol{A}$ in his next move marks the square labeled 2, then player $\boldsymbol{B}$ marks the square labeled 5. Player $\boldsymbol{B}$ wins as the square labeled 3 is left unmarked but cannot be covered with an $L$-figure. Finally, if player $\boldsymbol{A}$ in his next move marks one of the squares labeled 3 or 4, player $\boldsymbol{B}$ marks the other of these two squares. Player $\boldsymbol{B}$ wins as the square labeled 2 is left unmarked but cannot be covered with an $L$-figure. Since we have covered all possible cases, player $\boldsymbol{B}$ wins when $k=4$.

4

Combinatorics

math-word-problem

Incomplete

Yes

olympiads

C1. Consider a regular $2 n+1$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in 3 colors, such that every side is colored in exactly one color, and each color must be used at least once. Moreover, from every point in the plane external to $P$, at most 2 different colors on $P$ can be seen (ignore the vertices of $P$, we consider them colorless). Find the largest positive integer for which such a coloring is possible.

Solution. Answer: $n=1$ is clearly a solution, we can just color each side of the equilateral triangle in a different color, and the conditions are satisfied. We prove there is no larger $n$ that fulfills the requirements. Lemma 1. Given a regular $2 n+1$-gon in the plane, and a sequence of $n+1$ consecutive sides $s_{1}, s_{2}, \ldots, s_{n+1}$ there is an external point $Q$ in the plane, such that the color of each $s_{i}$ can be seen from $Q$, for $i=1,2, \ldots, n+1$. Proof. It is obvious that for a semi-circle $S$, there is a point $R$ in the plane far enough on the perpendicular bisector of the diameter of $S$ such that almost the entire semi-circle can be seen from $R$. Now, it is clear that looking at the circumscribed circle around the $2 n+1$-gon, there is a semi-circle $S$ such that each $s_{i}$ either has both endpoints on it, or has an endpoint that is on the semi-circle, and is not on the semicircle's end. So, take $Q$ to be a point in the plane from which almost all of $S$ can be seen, clearly, the color of each $s_{i}$ can be seen from $Q$. $\diamond$ Take $n \geq 2$, denote the sides $a_{1}, a_{2}, \ldots, a_{2 n+1}$ in that order, and suppose we have a coloring that satisfies the condition of the problem. Let's call the 3 colors red, green and blue. We must have 2 adjacent sides of different colors, say $a_{1}$ is red and $a_{2}$ is green. Then, by Lemma 1 : (i) We cannot have a blue side among $a_{1}, a_{2}, \ldots, a_{n+1}$. (ii) We cannot have a blue side among $a_{2}, a_{1}, a_{2 n+1}, \ldots, a_{n+3}$. We are required to have at least one blue side, and according to 1 ) and 2), that can only be $a_{n+2}$, so $a_{n+2}$ is blue. Now, applying Lemma 1 on the sequence of sides $a_{2}, a_{3}, \ldots, a_{n+2}$ we get that $a_{2}, a_{3}, \ldots, a_{n+1}$ are all green. Applying Lemma 1 on the sequence of sides $a_{1}, a_{2 n+1}, a_{2 n}, \ldots, a_{n+2}$ we get that $a_{2 n+1}, a_{2 n}, \ldots, a_{n+3}$ are all red. Therefore $a_{n+1}, a_{n+2}$ and $a_{n+3}$ are all of different colors, and for $n \geq 2$ they can all be seen from the same point according to Lemma 1 , so we have a contradiction.

1

Combinatorics

math-word-problem

Yes

Yes

olympiads

NT5. Find all positive integers $n$ such that there exists a prime number $p$, such that $$ p^{n}-(p-1)^{n} $$ is a power of 3 . Note. A power of 3 is a number of the form $3^{a}$ where $a$ is a positive integer.

Solution. Suppose that the positive integer $n$ is such that $$ p^{n}-(p-1)^{n}=3^{a} $$ for some prime $p$ and positive integer $a$. If $p=2$, then $2^{n}-1=3^{a}$ by $(1)$, whence $(-1)^{n}-1 \equiv 0(\bmod 3)$, so $n$ should be even. Setting $n=2 s$ we obtain $\left(2^{s}-1\right)\left(2^{s}+1\right)=3^{a}$. It follows that $2^{s}-1$ and $2^{s}+1$ are both powers of 3 , but since they are both odd, they are co-prime, and we have $2^{s}-1=1$, i.e. $s=1$ and $n=2$. If $p=3$, then (1) gives $3 \mid 2^{n}$, which is impossible. Let $p \geq 5$. Then it follows from (1) that we can not have $3 \mid p-1$. This means that $2^{n}-1 \equiv 0$ $(\bmod 3)$, so $n$ should be even, and let $n=2 k$. Then $$ p^{2 k}-(p-1)^{2 k}=3^{a} \Longleftrightarrow\left(p^{k}-(p-1)^{k}\right)\left(p^{k}+(p-1)^{k}\right)=3^{a} $$ If $d=\left(p^{k}-(p-1)^{k}, p^{k}+(p-1)^{k}\right)$, then $d \mid 2 p^{k}$. However, both numbers are powers of 3 , so $d=1$ and $p^{k}-(p-1)^{k}=1, p^{k}+(p-1)^{k}=3^{a}$. If $k=1$, then $n=2$ and we can take $p=5$. For $k \geq 2$ we have $1=p^{k}-(p-1)^{k} \geq p^{2}-(p-1)^{2}$ (this inequality is equivalent to $p^{2}\left(p^{k-2}-1\right) \geq(p-1)^{2}\left((p-1)^{k-2}-1\right)$, which is obviously true). Then $1 \geq p^{2}-(p-1)^{2}=2 p-1 \geq 9$, which is absurd. It follows that the only solution is $n=2$.

2

Number Theory

math-word-problem

Yes

Yes

olympiads

A5. Find the largest positive integer $n$ for which the inequality $$ \frac{a+b+c}{a b c+1}+\sqrt[n]{a b c} \leq \frac{5}{2} $$ holds for all $a, b, c \in[0,1]$. Here $\sqrt[1]{a b c}=a b c$.

Solution. Let $n_{\max }$ be the sought largest value of $n$, and let $E_{a, b, c}(n)=\frac{a+b+c}{a b c+1}+\sqrt[n]{a b c}$. Then $E_{a, b, c}(m)-E_{a, b, c}(n)=\sqrt[m]{a b c}-\sqrt[n]{a b c}$ and since $a . b c \leq 1$ we clearly have $E_{a, b, c}(m) \geq$ $E_{a, b, c}(n)$ for $m \geq n$. So if $E_{a, b, c}(n) \geq \frac{5}{2}$ for some choice of $a, b, c \in[0,1]$, it must be $n_{\max } \leq n$. We use this remark to determine the upper bound $n_{\max } \leq 3$ by plugging some particular values of $a, b, c$ into the given inequality as follow's: $$ \text { For }(a, b, c)=(1,1, c), c \in[0,1] \text {, inequality (1) implies } \frac{c+2}{c+1}+\sqrt[n]{c} \leq \frac{5}{2} \Leftrightarrow \frac{1}{c+1}+\sqrt[n]{c} \leq $$ $\frac{3}{2}$. Obviously, every $x \in[0 ; 1]$ is written as $\sqrt[n]{c}$ for some $c \in[0 ; 1]$. So the last inequality is equivalent to: $$ \begin{aligned} & \frac{1}{x^{n}+1}+x \leq \frac{3}{2} \Leftrightarrow 2+2 x^{n+1}+2 x \leq 3 x^{n}+3 \Leftrightarrow 3 x^{n}+1 \geq 2 x^{n+1}+2 x \\ \Leftrightarrow & 2 x^{n}(1-x)+(1-x)+(x-1)\left(x^{n-1}+\cdots+x\right) \geq 0 \\ \Leftrightarrow & (1-x)\left[2 x^{n}+1-\left(x^{n-1}+x^{n-2}+\ldots+x\right)\right] \geq 0, \forall x \in[0,1] \end{aligned} $$ For $n=4$, the left hand side of the above becomes $(1-x)\left(2 x^{4}+1-x^{3}-x^{2}-x\right)=$ $(1-x)(x-1)\left(2 x^{3}+x^{2}-1\right)=-(1-x)^{2}\left(2 x^{3}+x^{2}-1\right)$ which for $x=0.9$ is negative. Thus. $n_{\max } \leq 3$ as claimed. Now, we shall prove that for $n=3$ inequality (1) holds for all $a, b, c \in[0,1]$, and this would mean $n_{\max }=3$. We shall use the following Lemma: Lemma. For all $a, b, c \in[0 ; 1]: a+b+c \leq a b c+2$. Proof of the Lemma: The required result comes by adding the following two inequalities side by side $$ \begin{aligned} & 0 \leq(a-1)(b-1) \Leftrightarrow a+b \leq a b+1 \Leftrightarrow a+b-a b \leq 1 \\ & 0 \leq(a b-1)(c-1) \Leftrightarrow a b+c \leq a b c+1 \end{aligned} $$ Because of the Lemma, our inequality (1) for $n=3$ wrill be proved if the following weaker inequality is proved for all $a, b, c \in[0,1]$ : $$ \frac{a b c+2}{a b c+1}+\sqrt[3]{a b c} \leq \frac{5}{2} \Leftrightarrow \frac{1}{a b c+1}+\sqrt[3]{a b c} \leq \frac{3}{2} $$ Denoting $\sqrt[3]{a b c}=y \in[0 ; 1]$, this inequality becomes: $$ \begin{aligned} & \frac{1}{y^{3}+1}+y \leq \frac{3}{2} \Leftrightarrow 2+2 y^{4}+2 y \leq 3 y^{3}+3 \Leftrightarrow-2 y^{4}+3 y^{3}-2 y+1 \geq 0 \\ \Leftrightarrow & 2 y^{3}(1-y)+(y-1) y(y+1)+(1-y) \geq 0 \Leftrightarrow(1-y)\left(2 y^{3}+1-y^{2}-y\right) \geq 0 \end{aligned} $$ The last inequality is obvious because $1-y \geq 0$ and $2 y^{3}+1-y^{2}-y=y^{3}+(y-1)^{2}(y+1) \geq 0$. ## Geometry 2

3

Inequalities

math-word-problem

Yes

Yes

olympiads

G4. Let $A B C$ be an acute-angled triangle with circumcircle $\Gamma$, and let $O, H$ be the triangle's circumcenter and orthocenter respectively. Let also $A^{\prime}$ be the point where the angle bisector of angle $B A C$ meets $\Gamma$. If $A^{\prime} H=A H$, find the measure of angle $B A C$.  Figure 4: Exercise G4.

Solution. The segment $A A^{\prime}$ bisects $\angle O A H$ : if $\angle B C A=y$ (Figure 4), then $\angle B O A=$ $2 y$, and since $O A=O B$, it is $\angle O A B=\angle O B A=90^{\circ}-y$. Also since $A H \perp B C$, it is $\angle H A C=90^{\circ}-y=\angle O A B$ and the claim follows. Since $A . A^{\prime}$ bisects $\angle O A H$ and $A^{\prime} H=A H . O A^{\prime}=O A$, we have that the isosceles triangles $O A A^{\prime}, H A A^{\prime}$ are equal. Thus $$ A H=O A=R $$ where $R$ is the circumradius of triangle $A B C$. Call $\angle A C H=a$ and recall by the law of sines that $A H=2 R^{\prime} \sin a$, where $R^{\prime}$ is the circumradius of triangle $A H C$. Then (4) implies $$ R=2 R^{\prime} \sin a $$ But notice that $R=R^{\prime}$ because $\frac{A C}{\sin (A H C)}=2 R^{\prime}, \frac{A C}{\sin (A B C)}=2 R$ and $\sin (A H C)=$ $\sin \left(180^{\circ}-A B C\right)=\sin (A B C)$. So (5) gives $1=2 \sin a$, and $a$ as an acuite angle can only be $30^{\circ}$. Finally, $\angle B A C=90^{\circ}-a=60^{\circ}$. Remark. The steps in the above proof can be traced backwards making the converse also true, that is: If $\angle B A C=60^{\circ}$ then $A^{\prime} H=A H$. - G5. Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$ that touches them at $M$ and, $N$ respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.  Figure 5: Exercise G5.

60

Geometry

math-word-problem

Yes

Yes

olympiads

## A1 MLD Let $x, y, z$ be real numbers, satisfying the relations $$ \left\{\begin{array}{l} x \geq 20 \\ y \geq 40 \\ z \geq 1675 \\ x+y+z=2015 \end{array}\right. $$ Find the greatest value of the product $P=x \cdot y \cdot z$.

## Solution 1: By virtue of $z \geq 1675$ we have $$ y+z<2015 \Leftrightarrow y<2015-z \leq 2015-1675<1675 $$ It follows that $(1675-y) \cdot(1675-z) \leq 0 \Leftrightarrow y \cdot z \leq 1675 \cdot(y+z-1675)$. By using the inequality $u \cdot v \leq\left(\frac{u+v}{2}\right)^{2}$ for all real numbers $u, v$ we obtain $$ \begin{gathered} P=x \cdot y \cdot z \leq 1675 \cdot x \cdot(y+z-1675) \leq 1675 \cdot\left(\frac{x+y+z-1675}{2}\right)^{2}= \\ 1675 \cdot\left(\frac{2015-1675}{2}\right)^{2}=1675 \cdot 170^{2}=48407500 \end{gathered} $$ $$ \text { We have } P=x \cdot y \cdot z=48407500 \Leftrightarrow\left\{\begin{array} { l } { x + y + z = 2 0 1 5 , } \\ { z = 1 6 7 5 , } \\ { x = y + z - 1 6 7 5 } \end{array} \Leftrightarrow \left\{\begin{array}{l} x=170 \\ y=170 \\ z=1675 \end{array}\right.\right. $$ So, the greatest value of the product is $P=x \cdot y \cdot z=48407500$.

48407500

Algebra

math-word-problem

Yes

Yes

olympiads

## A2 ALB 3) If $x^{3}-3 \sqrt{3} x^{2}+9 x-3 \sqrt{3}-64=0$, find the value of $x^{6}-8 x^{5}+13 x^{4}-5 x^{3}+49 x^{2}-137 x+2015$.

Solution $x^{3}-3 \sqrt{3} x^{2}+9 x-3 \sqrt{3}-64=0 \Leftrightarrow(x-\sqrt{3})^{3}=64 \Leftrightarrow(x-\sqrt{3})=4 \Leftrightarrow x-4=\sqrt{3} \Leftrightarrow x^{2}-8 x+16=3 \Leftrightarrow$ $x^{2}-8 x+13=0$ $x^{6}-8 x^{5}+13 x^{4}-5 x^{3}+49 x^{2}-137 x+2015=\left(x^{2}-8 x+13\right)\left(x^{4}-5 x+9\right)+1898=0+1898=1898$

1898

Algebra

math-word-problem

Yes

Yes

olympiads

## A3 MNE Let $a, b, c$ be positive real numbers. Prove that $$ \frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2 $$

## Solution: Starting from the double expression on the left-hand side of given inequality, and applying twice the Arithmetic-Geometric mean inequality, we find that $$ \begin{aligned} 2 \frac{a}{b}+2 \sqrt{\frac{b}{c}}+2 \sqrt[3]{\frac{c}{a}} & =\frac{a}{b}+\left(\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt{\frac{b}{c}}\right)+2 \sqrt[3]{\frac{c}{a}} \\ & \geq \frac{a}{b}+3 \sqrt[3]{\frac{a}{b}} \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c}}+2 \sqrt[3]{\frac{c}{a}} \\ & =\frac{a}{b}+3 \sqrt[3]{\frac{a}{c}}+2 \sqrt[3]{\frac{c}{a}} \\ & \left.=\frac{a}{b}+\sqrt[3]{\frac{a}{c}}+2 \sqrt[3]{\frac{a}{c}}+\sqrt[3]{\frac{c}{a}}\right) \\ & \geq \frac{a}{b}+\sqrt[3]{\frac{a}{c}}+2 \cdot 2 \sqrt{\sqrt[3]{\frac{a}{c}} \sqrt[3]{\frac{c}{a}}} \\ & =\frac{a}{b}+\sqrt[3]{\frac{a}{c}}+4 \\ & >4 \end{aligned} $$ which yields the given inequality. (A4) GRE Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of $$ A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c} $$

4

Inequalities

proof

Yes

Yes

olympiads

NT1 SAU What is the greatest number of integers that can be selected from a set of 2015 consecutive numbers so that no sum of any two selected numbers is divisible by their difference?

## Solution: We take any two chosen numbers. If their difference is 1 , it is clear that their sum is divisible by their difference. If their difference is 2 , they will be of the same parity, and their sum is divisible by their difference. Therefore, the difference between any chosen numbers will be at least 3 . In other words, we can choose at most one number of any three consecutive numbers. This implies that we can choose at most 672 numbers. Now, we will show that we can choose 672 such numbers from any 2015 consecutive numbers. Suppose that these numbers are $a, a+1, \ldots, a+2014$. If $a$ is divisible by 3 , we can choose $a+1, a+4, \ldots, a+2014$. If $a$ is not divisible by 3 , we can choose $a, a+3, \ldots, a+$ 2013.

672

Number Theory

math-word-problem

Yes

Yes

olympiads

## C3 ALB Positive integers are put into the following table | 1 | 3 | 6 | 10 | 15 | 21 | 28 | 36 | | | | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | | 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 | | | | 4 | 8 | 13 | 19 | 26 | 34 | 43 | 53 | | | | 7 | 12 | 18 | 25 | 33 | 42 | | | | | | 11 | 17 | 24 | 32 | 41 | | | | | | | 16 | 23 | | | | | | | | | | $\ldots$ | | | | | | | | | | | $\ldots$ | | | | | | | | | | Find the number of the line and column where the number 2015 stays.

## Solution 1: We shall observe straights lines as on the next picture. We can call these lines diagonals. | 1 | $\sqrt{3}$ | 6 | 10 | 15 | 21 | 28 | 36 | | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 | | | 4 | 8 | 13 | 19 | 26 | 34 | 43 | 53 | | | | 12 | 18 | 25 | 33 | 42 | | | | | 11 | 17 | 24 | 32 | 41 | | | | |  On the first diagonal is number 1 . On the second diagonal are two numbers: 2 and 3 . On the 3rd diagonal are three numbers: 4,5 and 6 . On the $n$-th diagonal are $n$ numbers. These numbers are greater then $\frac{(n-1) n}{2}$ and not greater than $\frac{n(n+1)}{2}$ (see the next sentence!). On the first $n$ diagonals are $1+2+3+\ldots+n=\frac{n(n+1)}{2}$ numbers. If $m$ is in the $k$-th row $l$-th column and on the $n$-th diagonal, then it is $m=\frac{(n-1) n}{2}+l$ and $n+1=k+l$. So, $m=\frac{(k+l-2)(k+l-1)}{2}+l$. We have to find such numbers $n, k$ and $l$ for which: $$ \begin{gathered} \frac{(n-1) n}{2}<2015 \leq \frac{n(n+1)}{2} \\ n+1=k+l \\ 2015=\frac{(k+l-2)(k+l-1)}{2}+l \end{gathered} $$ (1), (2), (3) $\Rightarrow n^{2}-n<4030 \leq n^{2}+n \Rightarrow n=63, k+l=64,2015=\frac{(64-2)(64-1)}{2}+l \Rightarrow$ $t=2015-31 \cdot 63=62, k=64-62=2$ Therefore 2015 is located in the second row and 62 -th column.

2015

Number Theory

math-word-problem

Yes

Yes

olympiads

C2 Consider 50 points in the plane, no three of them belonging to the same line. The points have been colored into four colors. Prove that there are at least 130 scalene triangles whose vertices are colored in the same color.

## Solution Since $50=4 \cdot 12+2$, according to the pigeonhole principle we will have at least 13 points colored in the same color. We start with the: Lemma. Given $n>8$ points in the plane, no three of them collinear, then there are at least $\frac{n(n-1)(n-8)}{6}$ scalene triangles with vertices among the given points. Proof. There are $\frac{n(n-1)}{2}$ segments and $\frac{n(n-1)(n-2)}{6}$ triangles with vertices among the given points. We shall prove that there are at most $n(n-1)$ isosceles triangles. Indeed, for every segment $A B$ we can construct at most two isosceles triangles (if we have three $A B C, A B D$ and $A B E$, than $C, D, E$ will be collinear). Hence we have at least $$ \frac{n(n-1)(n-2)}{6}-n(n-1)=\frac{n(n-1)(n-8)}{6} \text { scalene triangles. } $$ For $n=13$ we have $\frac{13 \cdot 12 \cdot 5}{6}=130$, QED.

130

Combinatorics

proof

Yes

Yes

olympiads

G2 Let $A B C D$ be a convex quadrilateral with $\varangle D A C=\varangle B D C=36^{\circ}, \varangle C B D=18^{\circ}$ and $\varangle B A C=72^{\circ}$. If $P$ is the point of intersection of the diagonals $A C$ and $B D$, find the measure of $\varangle A P D$.

## Solution On the rays ( $D A$ and ( $B A$ we take points $E$ and $Z$, respectively, such that $A C=A E=$ $A Z$. Since $\varangle D E C=\frac{\varangle D A C}{2}=18^{\circ}=\varangle C B D$, the quadrilateral $D E B C$ is cyclic. Similarly, the quadrilateral $C B Z D$ is cyclic, because $\varangle A Z C=\frac{\varangle B A C}{2}=36^{\circ}=\varangle B D C$. Therefore the pentagon $B C D Z E$ is inscribed in the circle $k(A, A C)$. It gives $A C=A D$ and $\varangle A C D=\varangle A D C=\frac{180^{\circ}-36^{\circ}}{2}=72^{\circ}$, which gives $\varangle A D P=36^{\circ}$ and $\varangle A P D=108^{\circ}$. 

108

Geometry

math-word-problem

Yes

Yes

olympiads

## A1 For any real number a, let $\lfloor a\rfloor$ denote the greatest integer not exceeding a. In positive real numbers solve the following equation $$ n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor=2014 $$

Solution1. Obviously $n$ must be positive integer. Now note that $44^{2}=19362000$ than $2014=n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor>2000+44+12=2056$, a contradiction! So $1950 \leq n \leq 2000$, therefore $\lfloor\sqrt{n}\rfloor=44$ and $\lfloor\sqrt[3]{n}\rfloor=12$. Plugging that into the original equation we get: $$ n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor=n+44+12=2014 $$ From which we get $n=1956$, which is the only solution.

1956

Number Theory

math-word-problem

Yes

Yes

olympiads

## C2 In a country with $n$ cities, all direct airlines are two-way. There are $r>2014$ routes between pairs of different cities that include no more than one intermediate stop (the direction of each route matters). Find the least possible $n$ and the least possible $r$ for that value of $n$.

Solution. Denote by $X_{1}, X_{2}, \ldots X_{n}$ the cities in the country and let $X_{i}$ be connected to exactly $m_{i}$ other cities by direct two-way airline. Then $X_{i}$ is a final destination of $m_{i}$ direct routes and an intermediate stop of $m_{i}\left(m_{i}-1\right)$ non-direct routes. Thus $r=m_{1}^{2}+\ldots+m_{n}^{2}$. As each $m_{i}$ is at most $n-1$ and $13 \cdot 12^{2}<2014$, we deduce $n \geq 14$. Consider $n=14$. As each route appears in two opposite directions, $r$ is even, so $r \geq 2016$. We can achieve $r=2016$ by arranging the 14 cities uniformly on a circle connect (by direct two-way airlines) all of them, except the diametrically opposite pairs. This way, there are exactly $14 \cdot 12^{2}=2016$ routes.

2016

Combinatorics

math-word-problem

Yes

Yes

olympiads

## C4 Let $A=1 \cdot 4 \cdot 7 \cdot \ldots \cdot 2014$ be the product of the numbers less or equal to 2014 that give remainder 1 when divided by 3 . Find the last non-zero digit of $A$.

Solution. Grouping the elements of the product by ten we get: $$ \begin{aligned} & (30 k+1)(30 k+4)(30 k+7)(30 k+10)(30 k+13)(30 k+16) \\ & (30 k+19)(30 k+22)(30 k+25)(30 k+28)= \\ & =(30 k+1)(15 k+2)(30 k+7)(120 k+40)(30 k+13)(15 k+8) \\ & (30 k+19)(15 k+11)(120 k+100)(15 k+14) \end{aligned} $$ (We divide all even numbers not divisible by five, by two and multiply all numbers divisible by five with four.) We denote $P_{k}=(30 k+1)(15 k+2)(30 k+7)(30 k+13)(15 k+8)(30 k+19)(15 k+11)(15 k+14)$. For all the numbers not divisible by five, only the last digit affects the solution, since the power of two in the numbers divisible by five is greater than the power of five. Considering this, for even $k, P_{k}$ ends with the same digit as $1 \cdot 2 \cdot 7 \cdot 3 \cdot 8 \cdot 9 \cdot 1 \cdot 4$, i.e. six and for odd $k, P_{k}$ ends with the same digit as $1 \cdot 7 \cdot 7 \cdot 3 \cdot 3 \cdot 9 \cdot 6 \cdot 9$, i.e. six. Thus $P_{0} P_{1} \ldots P_{66}$ ends with six. If we remove one zero from the end of all numbers divisible with five, we get that the last nonzero digit of the given product is the same as the one from $6 \cdot 2011 \cdot 2014 \cdot 4 \cdot 10 \cdot 16 \cdot \ldots .796 \cdot 802$. Considering that $4 \cdot 6 \cdot 2 \cdot 8$ ends with four and removing one zero from every fifth number we get that the last nonzero digit is the same as in $4 \cdot 4^{26} \cdot 784 \cdot 796 \cdot 802 \cdot 1 \cdot 4 \cdot \ldots \cdot 76 \cdot 79$. Repeating the process we did for the starting sequence we conclude that the last nonzero number will be the same as in $2 \cdot 6 \cdot 6 \cdot 40 \cdot 100 \cdot 160 \cdot 220 \cdot 280 \cdot 61 \cdot 32 \cdot 67 \cdot 73 \cdot 38 \cdot 79$, which is two. Let $A B C$ be a triangle with $\measuredangle B=\measuredangle C=40^{\circ}$. The bisector of the $\measuredangle B$ meets $A C$ at the point $D$ . Prove that $\overline{B D}+\overline{D A}=\overline{B C}$.

2

Number Theory

math-word-problem

Yes

Yes

olympiads

A3. Let $A$ and $B$ be two non-empty subsets of $X=\{1,2, \ldots, 11\}$ with $A \cup B=X$. Let $P_{A}$ be the product of all elements of $A$ and let $P_{B}$ be the product of all elements of $B$. Find the minimum and maximum possible value of $P_{A}+P_{B}$ and find all possible equality cases.

Solution. For the maximum, we use the fact that $\left(P_{A}-1\right)\left(P_{B}-1\right) \geqslant 0$, to get that $P_{A}+P_{B} \leqslant P_{A} P_{B}+1=11!+1$. Equality holds if and only if $A=\{1\}$ or $B=\{1\}$. For the minimum observe, first that $P_{A} \cdot P_{B}=11!=c$. Without loss of generality let $P_{A} \leqslant P_{B}$. In this case $P_{A} \leqslant \sqrt{c}$. We write $P_{A}+P_{B}=P_{A}+\frac{c}{P_{A}}$ and consider the function $f(x)=x+\frac{c}{x}$ for $x \leqslant \sqrt{c}$. Since $$ f(x)-f(y)=x-y+\frac{c(y-x)}{y x}=\frac{(x-y)(x y-c)}{x y} $$ then $f$ is decreasing for $x \in(0, c]$. Since $x$ is an integer and cannot be equal with $\sqrt{c}$, the minimum is attained to the closest integer to $\sqrt{c}$. We have $\lfloor\sqrt{11!}\rfloor=\left\lfloor\sqrt{2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7 \cdot 11}\right\rfloor=\lfloor 720 \sqrt{77}\rfloor=6317$ and the closest integer which can be a product of elements of $X$ is $6300=2 \cdot 5 \cdot 7 \cdot 9 \cdot 10$. Therefore the minimum is $f(6300)=6300+6336=12636$ and it is achieved for example for $A=\{2,5,7,9,10\}, B=\{1,3,4,6,8,11\}$. Suppose now that there are different sets $A$ and $B$ such that $P_{A}+P_{B}=402$. Then the pairs of numbers $(6300,6336)$ and $\left(P_{A}, P_{B}\right)$ have the same sum and the same product, thus the equality case is unique for the numbers 6300 and 6336. It remains to find all possible subsets $A$ with product $6300=2^{2} \cdot 3^{2} \cdot 5^{2} \cdot 7$. It is immediate that $5,7,10 \in A$ and from here it is easy to see that all posibilities are $A=\{2,5,7,9,10\},\{1,2,5,7,9,10\},\{3,5,6,7,10\}$ and $\{1,3,5,6,7,10\}$.

12636

Combinatorics

math-word-problem

Yes

Yes

olympiads

C3. In a $5 \times 100$ table we have coloured black $n$ of its cells. Each of the 500 cells has at most two adjacent (by side) cells coloured black. Find the largest possible value of $n$.

Solution. If we colour all the cells along all edges of the board together with the entire middle row except the second and the last-but-one cell, the condition is satisfied and there are 302 black cells. The figure below exhibits this colouring for the $5 \times 8$ case.  We can cover the table by one fragment like the first one on the figure below, 24 fragments like the middle one, and one fragment like the third one.  In each fragment, among the cells with the same letter, there are at most two coloured black, so the total number of coloured cells is at most $(5+24 \cdot 6+1) \cdot 2+2=302$.

302

Combinatorics

math-word-problem

Yes

Yes

olympiads

C4. We have a group of $n$ kids. For each pair of kids, at least one has sent a message to the other one. For each kid $A$, among the kids to whom $A$ has sent a message, exactly $25 \%$ have sent a message to $A$. How many possible two-digit values of $n$ are there?

Solution. If the number of pairs of kids with two-way communication is $k$, then by the given condition the total number of messages is $4 k+4 k=8 k$. Thus the number of pairs of kids is $\frac{n(n-1)}{2}=7 k$. This is possible only if $n \equiv 0,1 \bmod 7$. - In order to obtain $n=7 m+1$, arrange the kids in a circle and let each kid send a message to the first $4 m$ kids to its right and hence receive a message from the first $4 m$ kids to its left. Thus there are exactly $m$ kids to which it has both sent and received messages. - In order to obtain $n=7 m$, let kid $X$ send no messages (and receive from every other kid). Arrange the remaining $7 m-1$ kids in a circle and let each kid on the circle send a message to the first $4 m-1$ kids to its right and hence receive a message from the first $4 m-1$ kids to its left. Thus there are exactly $m$ kids to which it has both sent and received messages. There are 26 two-digit numbers with remainder 0 or 1 modulo 7 . (All numbers of the form $7 m$ and $7 m+1$ with $2 \leqslant m \leqslant 14$.)

26

Combinatorics

math-word-problem

Yes

Yes

olympiads

C5. An economist and a statistician play a game on a calculator which does only one operation. The calculator displays only positive integers and it is used in the following way: Denote by $n$ an integer that is shown on the calculator. A person types an integer, $m$, chosen from the set $\{1,2, \ldots, 99\}$ of the first 99 positive integers, and if $m \%$ of the number $n$ is again a positive integer, then the calculator displays $m \%$ of $n$. Otherwise, the calculator shows an error message and this operation is not allowed. The game consists of doing alternatively these operations and the player that cannot do the operation looses. How many numbers from $\{1,2, \ldots, 2019\}$ guarantee the winning strategy for the statistician, who plays second? For example, if the calculator displays 1200, the economist can type 50 , giving the number 600 on the calculator, then the statistician can type 25 giving the number 150 . Now, for instance, the economist cannot type 75 as $75 \%$ of 150 is not a positive integer, but can choose 40 and the game continues until one of them cannot type an allowed number.

Solution. First of all, the game finishes because the number on the calculator always decreases. By picking $m \%$ of a positive integer $n$, players get the number $$ \frac{m \cdot n}{100}=\frac{m \cdot n}{2^{2} 5^{2}} $$ We see that at least one of the powers of 2 and 5 that divide $n$ decreases after one move, as $m$ is not allowed to be 100 , or a multiple of it. These prime divisors of $n$ are the only ones that can decrease, so we conclude that all the other prime factors of $n$ are not important for this game. Therefore, it is enough to consider numbers of the form $n=2^{k} 5^{\ell}$ where $k, \ell \in \mathbb{N}_{0}$, and to draw conclusions from these numbers. We will describe all possible changes of $k$ and $\ell$ in one move. Since $5^{3}>100$, then $\ell$ cannot increase, so all possible changes are from $\ell$ to $\ell+b$, where $b \in\{0,-1,-2\}$. For $k$, we note that $2^{6}=64$ is the biggest power of 2 less than 100 , so $k$ can be changed to $k+a$, where $a \in\{-2,-1,0,1,2,3,4\}$. But the changes of $k$ and $\ell$ are not independent. For example, if $\ell$ stays the same, then $m$ has to be divisible by 25 , giving only two possibilities for a change $(k, \ell) \rightarrow(k-2, \ell)$, when $m=25$ or $m=75$, or $(k, \ell) \rightarrow(k-1, \ell)$, when $m=50$. Similarly, if $\ell$ decreases by 1 , then $m$ is divisible exactly by 5 and then the different changes are given by $(k, \ell) \rightarrow(k+a, \ell-1)$, where $a \in\{-2,-1,0,1,2\}$, depending on the power of 2 that divides $m$ and it can be from $2^{0}$ to $2^{4}$. If $\ell$ decreases by 2 , then $m$ is not divisible by 5 , so it is enough to consider when $m$ is a power of two, giving changes $(k, \ell) \rightarrow(k+a, \ell-2)$, where $a \in\{-2,-1,0,1,2,3,4\}$. We have translated the starting game into another game with changing (the starting pair of non-negative integers) $(k, \ell)$ by moves described above and the player who cannot make the move looses, i.e. the player who manages to play the move $(k, \ell) \rightarrow(0,0)$ wins. We claim that the second player wins if and only if $3 \mid k$ and $3 \mid \ell$. We notice that all moves have their inverse modulo 3 , namely after the move $(k, \ell) \rightarrow$ $(k+a, \ell+b)$, the other player plays $(k+a, \ell+b) \rightarrow(k+a+c, \ell+b+d)$, where $$ (c, d) \in\{(0,-1),(0,-2),(-1,0),(-1,-1),(-1,-2),(-2,0),(-2,-1),(-2,-2)\} $$ is chosen such that $3 \mid a+c$ and $3 \mid b+d$. Such $(c, d)$ can be chosen as all possible residues different from $(0,0)$ modulo 3 are contained in the set above and there is no move that keeps $k$ and $\ell$ the same modulo 3 . If the starting numbers $(k, \ell)$ are divisible by 3 , then after the move of the first player at least one of $k$ and $\ell$ will not be divisible by 3 , and then the second player will play the move so that $k$ and $\ell$ become divisible by 3 again. In this way, the first player can never finish the game, so the second player wins. In all other cases, the first player will make such a move to make $k$ and $\ell$ divisible by 3 and then he becomes the second player in the game, and by previous reasoning, wins. The remaining part of the problem is to compute the number of positive integers $n \leqslant 2019$ which are winning for the second player. Those are the $n$ which are divisible by exactly $2^{3 k} 5^{3 \ell}, k, \ell \in \mathbb{N}_{0}$. Here, exact divisibility by $2^{3 k} 5^{3 \ell}$ in this context means that $2^{3 k} \| n$ and $5^{3 \ell} \| n$, even for $\ell=0$, or $k=0$. For example, if we say that $n$ is exactly divisible by 8 , it means that $8 \mid n, 16 \nmid n$ and $5 \nmid n$. We start by noting that for each ten consecutive numbers, exactly four of them coprime to 10 . Then we find the desired amount by dividing 2019 by numbers $2^{3 k} 5^{3 \ell}$ which are less than 2019 , and then computing the number of numbers no bigger than $\left\lfloor\frac{2019}{2^{3 k} 5^{3 \ell}}\right\rfloor$ which are coprime to 10 . First, there are $4 \cdot 201+4=808$ numbers (out of positive integers $n \leqslant 2019$ ) coprime to 10 . Then, there are $\left\lfloor\frac{2019}{8}\right\rfloor=252$ numbers divisible by 8 , and $25 \cdot 4+1=101$ among them are exactly divisible by 8 . There are $\left\lfloor\frac{2019}{64}\right\rfloor=31$ numbers divisible by 64 , giving $3 \cdot 4+1=13$ divisible exactly by 64 . And there are two numbers, 512 and $3 \cdot 512$, which are divisible by exactly 512 . Similarly, there are $\left\lfloor\frac{2019}{125}\right\rfloor=16$ numbers divisible by 125 , implying that $4+2=6$ of them are exactly divisible by 125 . Finally, there is only one number divisible by exactly 1000 , and this is 1000 itself. All other numbers that are divisible by exactly $2^{3 k} 5^{3 \ell}$ are greater than 2019. So, we obtain that $808+101+13+2+6+1=931$ numbers not bigger that 2019 are winning for the statistician.

931

Number Theory

math-word-problem

Yes

Incomplete

olympiads

G2 Let $A D, B F$ and $C E$ be the altitudes of $\triangle A B C$. A line passing through $D$ and parallel to $A B$ intersects the line $E F$ at the point $G$. If $H$ is the orthocenter of $\triangle A B C$, find the angle $\widehat{C G H}$.

## Solution 1 We can see easily that points $C, D, H, F$ lies on a circle of diameter $[C H]$. Take $\left\{F, G^{\prime}\right\}=\odot(C H F) \cap E F$. We have $\widehat{E F H}=\widehat{B A D}=\widehat{B C E}=\widehat{D F H}$ since the quadrilaterals $A E D C, A E H F, C D H F$ are cyclic. Hence $[F B$ is the bisector of $\widehat{E F D}$, so $H$ is the midpoint of the arc $D G^{\prime}$. It follows that $D G^{\prime} \perp C H$ since $[C H]$ is a diameter. Therefore $D G^{\prime} \| A B$ and $G \equiv G^{\prime}$. Finally $G$ lies on the circle $\odot(C F H)$, so $\widehat{H G C}=90^{\circ}$. 

90

Geometry

math-word-problem

Yes

Yes

olympiads

G3 Let $A B C$ be a triangle in which ( $B L$ is the angle bisector of $\widehat{A B C}(L \in A C), A H$ is an altitude of $\triangle A B C(H \in B C)$ and $M$ is the midpoint of the side $[A B]$. It is known that the midpoints of the segments $[B L]$ and $[M H]$ coincides. Determine the internal angles of triangle $\triangle A B C$.

## Solution Let $N$ be the intersection of the segments $[B L]$ and $[M H]$. Because $N$ is the midpoint of both segments $[B L]$ and $[M H]$, it follows that $B M L H$ is a parallelogram. This implies that $M L \| B C$ and $L H \| A B$ and hence, since $M$ is the midpoint of $[A B]$, the angle bisector [ $B L$ and the altitude $A H$ are also medians of $\triangle A B C$. This shows that $\triangle A B C$ is an equilateral one with all internal angles measuring $60^{\circ}$.

60

Geometry

math-word-problem

Yes

Yes

olympiads

NT2. Find all four digit numbers A such that $$ \frac{1}{3} A+2000=\frac{2}{3} \bar{A} $$ where $\bar{A}$ is the number with the same digits as $A$, but written in opposite order. (For example, $\overline{1234}=4321$.)

Solution. Let $A=1000 a+100 b+10 c+d$. Then we obtain the equality $$ \frac{1}{3}(1000 n+100 b+10 c+d)+2000=\frac{2}{3}(1000 d+100 c+10 b+a) $$ ol $$ 1999 d+190 c=80 b+998 a+6000 $$ It is clear that $d$ is even digit and $d>2$. So we have to investigate three cases: (i) $d=4$ : (ii) $d=6$; (iii) $d=8$. (i) If $d=4$. comparing the last digits in the upper equality we see that $a=2$ or $a=\overline{1}$. If $a=2$ then $19 c=80$, which is possible only when $40=c=0$. Hence the number $4=2004$ satisfies the condition. If $a=\overline{7}$ then $19 c-8 b=490$, which is impossible. (ii) If $d=6$ then $190 c+5994=80 b+998$ r. Comparing the last digits we obtain tilat. $a=3$ or $a=8$. If $a=3$ then $80 b+998 a599+190 c$. (iii) If $d=8$ then $190 c+9992=80 b+998 a$. Now $80 b+998 a \leq 80 \cdot 9+998 \cdot 9=9702<$ $9992+190 c$ Hence we have the only solution $4=2004$.

2004

Number Theory

math-word-problem

Yes

Incomplete

olympiads

C2. Given $m \times n$ table, each cell signed with "-". The following operations are (i) to change all the signs in entire row to the opposite, i. e. every "-" to "+", and every "+" to "-"; (ii) to change all the signs in entire column to the opposite, i. e. every "-" to "+" and every "+" to " -". (a) Prove that if $m=n=100$, using the above operations one can not obtain 2004 signs "t". (b) If $m=1004$, find the least $n>100$ for which 2004 signs " + " can be obtained.

Solution. If we apply (i) to $l$ rows and (ii) to $k$ columns we obtain $(m-k) l+(n-l) k$ (a) We have equation $(100-k) l+(100-l) k=2004$, or $100 l+100 k-2 l k=2004$, le $$ 50 l+50 k-1 k=1002 $$ Rewrite the lasc equation as $$ (50-l)(50-h)=2.500-100.2=1498 $$ Since $1498=2 \cdot 7 \cdot 107$, this equation has no solitions in natural numbers. (b) Let $n=101$. Then we have $$ (100-k) l+(101-l) k=2004 $$ OI $$ 100 l+101 k-2 l k=2004 $$ l.e. $$ 101 k=2004-100 l+2 l k \div 2(1002-50 l+l k) $$ Hence $s=2 t$ and we have $101 t=501-25 l+2 l t$. From here we have $$ t=\frac{501-25 l}{101-2 l}=4+\frac{97-17 l}{101-2 l} $$ Since $t$ is natural number and $97-17 l<101-2 l$, this is a contradiction, Hence $n \neq 101$. Let $n=1.02$. Then we have $$ (100-k) l+(102-l) k=2004 $$ or $$ 100 l+102 k-2 l k=2004 $$ $$ 50 l+51 k-l k=1002 $$ Rewrite the last equation as $$ (51-l)(50-k)=25.50-1002=1.548 $$ Since $145 S=2 \cdot 2 \cdot 3 \cdot 3 \cdot 43$ we have $51-l=36$ and $50-k=43$. From here obtain $l=15$ and $k=7$. Indeed, $$ (100-\bar{\imath}) \cdot 15+(102-1.5) \cdot \overline{7}=93 \cdot 15+87 \cdot 7=1395+609=2004 $$ Hence, the least $n$ is 102 .

102

Combinatorics

proof

Yes

Yes

olympiads

Problem A2. Determine all four digit numbers $\overline{a b c d}$ such that $$ a(a+b+c+d)\left(a^{2}+b^{2}+c^{2}+d^{2}\right)\left(a^{6}+2 b^{6}+3 c^{6}+4 d^{6}\right)=\overline{a b c d} $$

Solution. From $\overline{a b c d}\overline{1 b c d}=(1+b+c+d)\left(1+b^{2}+c^{2}+d^{2}\right)\left(1+2 b^{6}+3 c^{6}+4 d^{6}\right) \geq$ $(b+1)\left(b^{2}+1\right)\left(2 b^{6}+1\right)$, so $b \leq 2$. Similarly one gets $c\overline{2 b c d}=2(2+b+c+d)\left(4+b^{2}+c^{2}+d^{2}\right)\left(64+2 b^{6}+3 c^{6}+4 d^{6}\right) \geq$ $2(b+2)\left(b^{2}+4\right)\left(2 b^{6}+64\right)$, imposing $b \leq 1$. In the same way one proves $c<2$ and $d<2$. By direct check, we find out that 2010 is the only solution.

2010

Number Theory

math-word-problem

Yes

Yes

olympiads

Problem G2. Consider a triangle $A B C$ and let $M$ be the midpoint of the side $B C$. Suppose $\angle M A C=\angle A B C$ and $\angle B A M=105^{\circ}$. Find the measure of $\angle A B C$.

Solution. The angle measure is $30^{\circ}$.  Let $O$ be the circumcenter of the triangle $A B M$. From $\angle B A M=105^{\circ}$ follows $\angle M B O=15^{\circ}$. Let $M^{\prime}, C^{\prime}$ be the projections of points $M, C$ onto the line $B O$. Since $\angle M B O=15^{\circ}$, then $\angle M O M^{\prime}=30^{\circ}$ and consequently $M M^{\prime}=\frac{M O}{2}$. On the other hand, $M M^{\prime}$ joins the midpoints of two sides of the triangle $B C C^{\prime}$, which implies $C C^{\prime}=M O=A O$. The relation $\angle M A C=\angle A B C$ implies $C A$ tangent to $\omega$, hence $A O \perp A C$. It follows that $\triangle A C O \equiv \triangle O C C^{\prime}$, and furthermore $O B \| A C$. Therefore $\angle A O M=\angle A O M^{\prime}-\angle M O M^{\prime}=90^{\circ}-30^{\circ}=60^{\circ}$ and $\angle A B M=$ $\frac{\angle A O M}{2}=30^{\circ}$.

30

Geometry

math-word-problem

Yes

Yes

olympiads

Problem N2. Find all positive integers $n$ such that $36^{n}-6$ is a product of two or more consecutive positive integers.

Solution. Answer: $n=1$. Among each four consecutive integers there is a multiple of 4 . As $36^{n}-6$ is not a multiple of 4 , it must be the product of two or three consecutive positive integers. Case I. If $36^{n}-6=x(x+1)$ (all letters here and below denote positive integers), then $4 \cdot 36^{n}-23=(2 x+1)^{2}$, whence $\left(2 \cdot 6^{n}+2 x+1\right)\left(2 \cdot 6^{n}-2 x-1\right)=23$. As 23 is prime, this leads to $2 \cdot 6^{n}+2 x+1=23,2 \cdot 6^{n}-2 x-1=1$. Subtracting these yields $4 x+2=22, x=5, n=1$, which is a solution to the problem. Case II. If $36^{n}-6=(y-1) y(y+1)$, then $$ 36^{n}=y^{3}-y+6=\left(y^{3}+8\right)-(y+2)=(y+2)\left(y^{2}-2 y+3\right) $$ Thus each of $y+2$ and $y^{2}-2 y+3$ can have only 2 and 3 as prime factors, so the same is true for their GCD. This, combined with the identity $y^{2}-2 y+3=$ $(y+2)(y-4)+11$ yields $\operatorname{GCD}\left(y+2 ; y^{2}-2 y+3\right)=1$. Now $y+2<y^{2}-2 y+3$ and the latter number is odd, so $y+2=4^{n}, y^{2}-2 y+3=9^{n}$. The former identity implies $y$ is even and now by the latter one $9^{n} \equiv 3(\bmod 4)$, while in fact $9^{n} \equiv 1(\bmod 4)$ - a contradiction. So, in this case there is no such $n$.

1

Number Theory

math-word-problem

Yes

Yes

olympiads

A1. Let $a, b, c$ be positive real numbers such that $a b c=8$. Prove that $$ \frac{a b+4}{a+2}+\frac{b c+4}{b+2}+\frac{c a+4}{c+2} \geq 6 $$

Solution. We have $a b+4=\frac{8}{c}+4=\frac{4(c+2)}{c}$ and similarly $b c+4=\frac{4(a+2)}{a}$ and $c a+4=\frac{4(b+2)}{b}$. It follows that $$ (a b+4)(b c+4)(c a+4)=\frac{64}{a b c}(a+2)(b+2)(c+2)=8(a+2)(b+2)(c+2) $$ so that $$ \frac{(a b+4)(b c+4)(c a+4)}{(a+2)(b+2)(c+2)}=8 $$ Applying AM-GM, we conclude: $$ \frac{a b+4}{a+2}+\frac{b c+4}{b+2}+\frac{c a+4}{c+2} \geq 3 \cdot \sqrt[3]{\frac{(a b+4)(b c+4)(c a+4)}{(a+2)(b+2)(c+2)}}=6 $$ Alternatively, we can write LHS as $$ \frac{b c(a b+4)}{2(b c+4)}+\frac{a c(b c+4)}{2(a c+4)}+\frac{a b(c a+4)}{2(a b+4)} $$ and then apply AM-GM.

6

Inequalities

proof

Yes

Yes

olympiads

A3. Determine the number of pairs of integers $(m, n)$ such that $$ \sqrt{n+\sqrt{2016}}+\sqrt{m-\sqrt{2016}} \in \mathbb{Q} $$

Solution. Let $r=\sqrt{n+\sqrt{2016}}+\sqrt{m-\sqrt{2016}}$. Then $$ n+m+2 \sqrt{n+\sqrt{2016}} \cdot \sqrt{m-\sqrt{2016}}=r^{2} $$ and $$ (m-n) \sqrt{2106}=\frac{1}{4}\left(r^{2}-m-n\right)^{2}-m n+2016 \in \mathbb{Q} $$ Since $\sqrt{2016} \notin \mathbb{Q}$, it follows that $m=n$. Then $$ \sqrt{n^{2}-2016}=\frac{1}{2}\left(r^{2}-2 n\right) \in \mathbb{Q} $$ Hence, there is some nonnegative integer $p$ such that $n^{2}-2016=p^{2}$ and (1) becomes $2 n+2 p=r^{2}$. It follows that $2(n+p)=r^{2}$ is the square of a rational and also an integer, hence a perfect square. On the other hand, $2016=(n-p)(n+p)$ and $n+p$ is a divisor of 2016, larger than $\sqrt{2016}$. Since $n+p$ is even, so is also $n-p$, and $r^{2}=2(n+p)$ is a divisor of $2016=2^{5} \cdot 3^{2} \cdot 7$, larger than $2 \sqrt{2016}>88$. The only possibility is $r^{2}=2^{4} \cdot 3^{2}=12^{2}$. Hence, $n+p=72$ and $n-p=28$, and we conclude that $n=m=50$. Thus, there is only one such pair.

1

Algebra

math-word-problem

Yes

Yes

olympiads

C1. Let $S_{n}$ be the sum of reciprocal values of non-zero digits of all positive integers up to (and including) $n$. For instance, $S_{13}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{2}+\frac{1}{1}+\frac{1}{3}$. Find the least positive integer $k$ making the number $k!\cdot S_{2016}$ an integer.

## Solution. We will first calculate $S_{999}$, then $S_{1999}-S_{999}$, and then $S_{2016}-S_{1999}$. Writing the integers from 1 to 999 as 001 to 999, adding eventually also 000 (since 0 digits actually do not matter), each digit appears exactly 100 times in each position(as unit, ten, or hundred). Hence $$ S_{999}=300 \cdot\left(\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{9}\right) $$ For the numbers in the interval $1000 \rightarrow 1999$, compared to $0 \rightarrow 999$, there are precisely 1000 more digits 1 . We get $$ S_{1999}-S_{999}=1000+S_{999} \Longrightarrow S_{1999}=1000+600 \cdot\left(\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{9}\right) $$ Finally, in the interval $2000 \rightarrow 2016$, the digit 1 appears twice as unit and seven times as a ten, the digit 2 twice as a unit and 17 times as a thousand, the digits $3,4,5$, and 6 each appear exactly twice as units, and the digits 7,8 , and 9 each appear exactly once as a unit. Hence $$ S_{2016}-S_{1999}=9 \cdot 1+19 \cdot \frac{1}{2}+2 \cdot\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)+1 \cdot\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\right) $$ In the end, we get $$ \begin{aligned} S_{2016} & =1609 \cdot 1+619 \cdot \frac{1}{2}+602 \cdot\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)+601 \cdot\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\right) \\ & =m+\frac{1}{2}+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+\frac{2}{6}+\frac{6}{7}+\frac{1}{8}+\frac{7}{9}=n+\frac{p}{2^{3} \cdot 3^{2} \cdot 5 \cdot 7} \end{aligned} $$ where $m, n$, and $p$ are positive integers, $p$ coprime to $2^{3} \cdot 3^{2} \cdot 5 \cdot 7$. Then $k!\cdot S_{2016}$ is an integer precisely when $k$ ! is a multiple of $2^{3} \cdot 3^{2} \cdot 5 \cdot 7$. Since $7 \mid k!$, it follows that $k \geq 7$. Also, $7!=2^{4} \cdot 3^{2} \cdot 5 \cdot 7$, implying that the least $k$ satisfying $k!\cdot S_{2016} \in \mathbb{Z}$ is $k=7$.

7

Number Theory

math-word-problem

Yes

Yes

olympiads

C2. The natural numbers from 1 to 50 are written down on the blackboard. At least how many of them should be deleted, in order that the sum of any two of the remaining numbers is not a prime?

Solution. Notice that if the odd, respectively even, numbers are all deleted, then the sum of any two remaining numbers is even and exceeds 2 , so it is certainly not a prime. We prove that 25 is the minimal number of deleted numbers. To this end, we group the positive integers from 1 to 50 in 25 pairs, such that the sum of the numbers within each pair is a prime: $$ \begin{aligned} & (1,2),(3,4),(5,6),(7,10),(8,9),(11,12),(13,16),(14,15),(17,20) \\ & (18,19),(21,22),(23,24),(25,28),(26,27),(29,30),(31,36),(32,35) \\ & (33,34),(37,42),(38,41),(39,40),(43,46),(44,45),(47,50),(48,49) \end{aligned} $$ Since at least one number from each pair has to be deleted, the minimal number is 25 .

25

Number Theory

math-word-problem

Yes

Yes

olympiads

C3. Consider any four pairwise distinct real numbers and write one of these numbers in each cell of a $5 \times 5$ array so that each number occurs exactly once in every $2 \times 2$ subarray. The sum over all entries of the array is called the total sum of that array. Determine the maximum number of distinct total sums that may be obtained in this way.

Solution. We will prove that the maximum number of total sums is 60 . The proof is based on the following claim. Claim. Either each row contains exactly two of the numbers, or each column contains exactly two of the numbers. Proof of the Claim. Indeed, let $R$ be a row containing at least three of the numbers. Then, in row $R$ we can find three of the numbers in consecutive positions, let $x, y, z$ be the numbers in consecutive positions(where $\{x, y, s, z\}=\{a, b, c, d\}$ ). Due to our hypothesis that in every $2 \times 2$ subarray each number is used exactly once, in the row above $\mathrm{R}$ (if there is such a row), precisely above the numbers $x, y, z$ will be the numbers $z, t, x$ in this order. And above them will be the numbers $x, y, z$ in this order. The same happens in the rows below $R$ (see at the following figure). $$ \left(\begin{array}{lllll} \bullet & x & y & z & \bullet \\ \bullet & z & t & x & \bullet \\ \bullet & x & y & z & \bullet \\ \bullet & z & t & x & \bullet \\ \bullet & x & y & z & \bullet \end{array}\right) $$ Completing all the array, it easily follows that each column contains exactly two of the numbers and our claim has been proven. Rotating the matrix (if it is necessary), we may assume that each row contains exactly two of the numbers. If we forget the first row and column from the array, we obtain a $4 \times 4$ array, that can be divided into four $2 \times 2$ subarrays, containing thus each number exactly four times, with a total sum of $4(a+b+c+d)$. It suffices to find how many different ways are there to put the numbers in the first row $R_{1}$ and the first column $C_{1}$. Denoting by $a_{1}, b_{1}, c_{1}, d_{1}$ the number of appearances of $a, b, c$, and respectively $d$ in $R_{1}$ and $C_{1}$, the total sum of the numbers in the entire $5 \times 5$ array will be $$ S=4(a+b+c+d)+a_{1} \cdot a+b_{1} \cdot b+c_{1} \cdot c+d_{1} \cdot d $$ If the first, the third and the fifth row contain the numbers $x, y$, with $x$ denoting the number at the entry $(1,1)$, then the second and the fourth row will contain only the numbers $z, t$, with $z$ denoting the number at the entry $(2,1)$. Then $x_{1}+y_{1}=7$ and $x_{1} \geqslant 3$, $y_{1} \geqslant 2, z_{1}+t_{1}=2$, and $z_{1} \geqslant t_{1}$. Then $\left\{x_{1}, y_{1}\right\}=\{5,2\}$ or $\left\{x_{1}, y_{1}\right\}=\{4,3\}$, respectively $\left\{z_{1}, t_{1}\right\}=\{2,0\}$ or $\left\{z_{1}, t_{1}\right\}=\{1,1\}$. Then $\left(a_{1}, b_{1}, c_{1}, d_{1}\right)$ is obtained by permuting one of the following quadruples: $$ (5,2,2,0),(5,2,1,1),(4,3,2,0),(4,3,1,1) $$ There are a total of $\frac{4!}{2!}=12$ permutations of $(5,2,2,0)$, also 12 permutations of $(5,2,1,1)$, 24 permutations of $(4,3,2,0)$ and finally, there are 12 permutations of $(4,3,1,1)$. Hence, there are at most 60 different possible total sums. We can obtain indeed each of these 60 combinations: take three rows ababa alternating with two rows $c d c d c$ to get $(5,2,2,0)$; take three rows ababa alternating with one row $c d c d c$ and a row $(d c d c d)$ to get $(5,2,1,1)$; take three rows $a b a b c$ alternating with two rows $c d c d a$ to get $(4,3,2,0)$; take three rows abcda alternating with two rows $c d a b c$ to get $(4,3,1,1)$. By choosing for example $a=10^{3}, b=10^{2}, c=10, d=1$, we can make all these sums different. Hence, 60 is indeed the maximum possible number of different sums.

60

Combinatorics

math-word-problem

Yes

Yes

olympiads

N1. Determine the largest positive integer $n$ that divides $p^{6}-1$ for all primes $p>7$.

Solution. Note that $$ p^{6}-1=(p-1)(p+1)\left(p^{2}-p+1\right)\left(p^{2}+p+1\right) $$ For $p=11$ we have $$ p^{6}-1=1771560=2^{3} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 19 \cdot 37 $$ For $p=13$ we have $$ p^{6}-1=2^{3} \cdot 3^{2} \cdot 7 \cdot 61 \cdot 157 $$ From the last two calculations we find evidence to try showing that $p^{6}-1$ is divisible by $2^{3} \cdot 3^{2} \cdot 7=504$ and this would be the largest positive integer that divides $p^{6}-1$ for all primes greater than 7 . By Fermat's theorem, $7 \mid p^{6}-1$. Next, since $p$ is odd, $8 \mid p^{2}-1=(p-1)(p+1)$, hence $8 \mid p^{6}-1$. It remains to show that $9 \mid p^{6}-1$. Any prime number $p, p>3$ is 1 or -1 modulo 3 . In the first case both $p-1$ and $p^{2}+p+1$ are divisible by 3 , and in the second case, both $p+1$ and $p^{2}-p+1$ are divisible by 3 . Consequently, the required number is indeed 504

504

Number Theory

math-word-problem

Yes

Yes

olympiads

N2. Find the maximum number of natural numbers $x_{1}, x_{2}, \ldots, x_{m}$ satisfying the conditions: a) No $x_{i}-x_{j}, 1 \leq i<j \leq m$ is divisible by 11 ; and b) The sum $x_{2} x_{3} \ldots x_{m}+x_{1} x_{3} \ldots x_{m}+\cdots+x_{1} x_{2} \ldots x_{m-1}$ is divisible by 11 .

Solution. The required maximum is 10 . According to a), the numbers $x_{i}, 1 \leq i \leq m$, are all different $(\bmod 11)$ (1) Hence, the number of natural numbers satisfying the conditions is at most 11. If $x_{j} \equiv 0(\bmod 11)$ for some $j$, then $$ x_{2} x_{3} \ldots x_{m}+x_{1} x_{3} \ldots x_{m}+\cdots+x_{1} x_{2} \ldots x_{m-1} \equiv x_{1} \ldots x_{j-1} x_{j+1} \ldots x_{m} \quad(\bmod 11) $$ which would lead to $x_{i} \equiv 0(\bmod 11)$ for some $i \neq j$, contradicting (1). We now prove that 10 is indeed the required maximum. Consider $x_{i}=i$, for all $i \in\{1,2, \ldots, 10\}$. The products $2 \cdot 3 \cdots \cdot 10,1 \cdot 3 \cdots \cdots 10, \ldots$, $1 \cdot 2 \cdots \cdot 9$ are all different $(\bmod 11)$, and so $$ 2 \cdot 3 \cdots \cdots 10+1 \cdot 3 \cdots \cdots 10+\cdots+1 \cdot 2 \cdots \cdot 9 \equiv 1+2+\cdots+10 \quad(\bmod 11) $$ and condition b) is satisfied, since $1+2+\cdots+10=55=5 \cdot 11$.

10

Number Theory

math-word-problem

Yes

Yes

olympiads

N5. Determine all four-digit numbers $\overline{a b c d}$ such that $$ (a+b)(a+c)(a+d)(b+c)(b+d)(c+d)=\overline{a b c d} $$

Solution. Depending on the parity of $a, b, c, d$, at least two of the factors $(a+b),(a+c)$, $(a+d),(b+c),(b+d),(c+d)$ are even, so that $4 \mid \overline{a b c d}$. We claim that $3 \mid \overline{a b c d}$. Assume $a+b+c+d \equiv 2(\bmod 3)$. Then $x+y \equiv 1(\bmod 3)$, for all distinct $x, y \in\{a, b, c, d\}$. But then the left hand side in the above equality is congruent to $1(\bmod 3)$ and the right hand side congruent to $2(\bmod 3)$, contradiction. Assume $a+b+c+d \equiv 1(\bmod 3)$. Then $x+y \equiv 2(\bmod 3)$, for all distinct $x, y \in\{a, b, c, d\}$, and $x \equiv 1(\bmod 3)$, for all $x, y \in\{a, b, c, d\}$. Hence, $a, b, c, d \in\{1,4,7\}$, and since $4 \mid \overline{a b c d}$, we have $c=d=4$. Therefore, $8 \mid \overline{a b 44}$, and since at least one more factor is even, it follows that $16 \overline{a b 44}$. Then $b \neq 4$, and the only possibilities are $b=1$, implying $a=4$, which is impossible because 4144 is not divisible by $5=1+4$, or $b=7$, implying $11 \mid \overline{a 744}$, hence $a=7$, which is also impossible because 7744 is not divisible by $14=7+7$. We conclude that $3 \mid \overline{a b c d}$, hence also $3 \mid a+b+c+d$. Then at least one factor $x+y$ of $(a+b),(a+c),(a+d),(b+c),(b+d),(c+d)$ is a multiple of 3 , implying that also $3 \mid a+b+c+d-x-y$, so $9 \mid \overline{a b c d}$. Then $9 \mid a+b+c+d$, and $a+b+c+d \in\{9,18,27,36\}$. Using the inequality $x y \geq x+y-1$, valid for all $x, y \in \mathbb{N}^{*}$, if $a+b+c+d \in\{27,36\}$, then $$ \overline{a b c d}=(a+b)(a+c)(a+d)(b+c)(b+d)(c+d) \geq 26^{3}>10^{4} $$ which is impossible. Using the inequality $x y \geq 2(x+y)-4$ for all $x, y \geq 2$, if $a+b+c+d=18$ and all two-digit sums are greater than 1 , then $\overline{a b c d} \geq 32^{3}>10^{4}$. Hence, if $a+b+c+d=18$, some two-digit sum must be 1 , hence the complementary sum will be 17 , and the digits are $\{a, b, c, d\}=\{0,1,8,9\}$. But then $\overline{a b c d}=1 \cdot 17 \cdot 8 \cdot 9^{2} \cdot 10>10^{4}$. We conclude that $a+b+c+d=9$. Then among $a, b, c, d$ there are either three odd or three even numbers, and $8 \mid \overline{a b c d}$. If three of the digits are odd, then $d$ is even and since $c$ is odd, divisibility by 8 implies that $d \in\{2,6\}$. If $d=6$, then $a=b=c=1$. But 1116 is not divisible by 7 , so this is not a solution. If $d=2$, then $a, b, c$ are either $1,1,5$ or $1,3,3$ in some order. In the first case $2 \cdot 6^{2} \cdot 3^{2} \cdot 7=4536 \neq \overline{a b c d}$. The second case cannot hold because the resulting number is not a multiple of 5 . Hence, there has to be one odd and three even digits. At least one of the two-digits sums of even digits is a multiple of 4 , and since there cannot be two zero digits, we have either $x+y=4$ and $z+t=5$, or $x+y=8$ and $z+t=1$ for some ordering $x, y, z, t$ of $a, b, c, d$. In the first case we have $d=0$ and the digits are $0,1,4,4$, or $0,2,3,4$, or $0,2,2,5$. None of these is a solution because $1 \cdot 4^{2} \cdot 5^{2} \cdot 8=3200,2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7=5040$ and $2^{2} \cdot 5 \cdot 4 \cdot 7^{2}=3920$. In the second case two of the digits are 0 and 1 , and the other two have to be either 4 and 4 , or 2 and 6 . We already know that the first possibility fails. For the second, we get $$ (0+1) \cdot(0+2) \cdot(0+6) \cdot(1+2) \cdot(1+6) \cdot(2+6)=2016 $$ and $\overline{a b c d}=2016$ is the only solution.

2016

Number Theory

math-word-problem

Yes

Yes

olympiads

NT 3. Find the largest integer $k(k \geq 2)$, for which there exists an integer $n(n \geq k)$ such that from any collection of $n$ consecutive positive integers one can always choose $k$ numbers, which verify the following conditions: 1. each chosen number is not divisible by 6 , by 7 and by 8 ; 2. the positive difference of any two different chosen numbers is not divisible by at least one of the numbers 6,7 or 8 .

Solution. An integer is divisible by 6,7 and 8 if and only if it is divisible by their Least Common Multiple, which equals $6 \times 7 \times 4=168$. Let $n$ be a positive integer and let $A$ be an arbitrary set of $n$ consecutive positive integers. Replace each number $a_{i}$ from $A$ with its remainder $r_{i}$ ( mod 168). The number $a_{i}$ is divisible by 6 ( 7 or 8 ) if and only if its remainder $r_{i}$ is divisible by 6 (respectively 7 or 8 ). The difference $\left|a_{i}-a_{j}\right|$ is divisible by 168 if and only if their remainders $r_{i}=r_{j}$. Choosing $k$ numbers from the initial set $A$, which verify the required conditions, is the same as choosing $k$ their remainders ( mod 168) such that: 1. each chosen remainder is not divisible by 6,7 and 8 ; 2. all chosen remainders are different. Suppose we have chosen $k$ numbers from $A$, which verify the conditions. Therefore, all remainders are different and $k \leq 168$ (otherwise, there would be two equal remainders). Denote by $B=\{0,1,2,3, \ldots, 167\}$ the set of all possible remainders ( $\bmod 168)$ and by $B_{m}$ the subset of all elements of $B$, which are divisible by $m$. Compute the number of elements of the following subsets: $$ \begin{gathered} \left|B_{6}\right|=168: 6=28, \quad\left|B_{7}\right|=168: 7=24, \quad\left|B_{8}\right|=168: 8=21 \\ \left|B_{6} \cap B_{7}\right|=\left|B_{42}\right|=168: 42=4, \quad\left|B_{6} \cap B_{8}\right|=\left|B_{24}\right|=168: 24=7 \\ \left|B_{7} \cap B_{8}\right|=\left|B_{56}\right|=168: 56=3, \quad\left|B_{6} \cap B_{7} \cap B_{8}\right|=\left|B_{168}\right|=1 \end{gathered} $$ Denote by $D=B_{6} \cup B_{7} \cup B_{8}$, the subset of all elements of $B$, which are divisible by at least one of the numbers 6,7 or 8 . By the Inclusion-Exclusion principle we got $$ \begin{gathered} |D|=\left|B_{6}\right|+\left|B_{7}\right|+\left|B_{8}\right|-\left(\left|B_{6} \cap B_{7}\right|+\left|B_{6} \cap B_{8}\right|+\left|B_{7} \cap B_{8}\right|\right)+\left|B_{6} \cap B_{7} \cap B_{8}\right|= \\ 28+24+21-(4+7+3)+1=60 . \end{gathered} $$ Each chosen remainder belongs to the subset $B \backslash D$, since it is not divisible by 6,7 and 8 . Hence, $k \leq|B \backslash D|=168-60=108$. Let us show that the greatest possible value is $k=108$. Consider $n=168$. Given any collection $A$ of 168 consecutive positive integers, replace each number with its remainder ( $\bmod 168$ ). Choose from these remainders 108 numbers, which constitute the set $B \backslash D$. Finally, take 108 numbers from the initial set $A$, having exactly these remainders. These $k=108$ numbers verify the required conditions.

108

Number Theory

math-word-problem

Yes

Yes

olympiads

A 2. Find the maximum positive integer $k$ such that for any positive integers $m, n$ such that $m^{3}+n^{3}>$ $(m+n)^{2}$, we have $$ m^{3}+n^{3} \geq(m+n)^{2}+k $$

Solution. We see that for $m=3$ and $n=2$ we have $m^{3}+n^{3}>(m+n)^{2}$, thus $$ 3^{3}+2^{3} \geq(3+2)^{2}+k \Rightarrow k \leq 10 $$ We will show that $k=10$ is the desired maximum. In other words, we have to prove that $$ m^{3}+n^{3} \geq(m+n)^{2}+10 $$ The last inequality is equivalent to $$ (m+n)\left(m^{2}+n^{2}-m n-m-n\right) \geq 10 $$ If $m+n=2$ or $m+n=3$, then $(m, n)=(1,1),(1,2),(2,1)$ and we can check that none of them satisfies the condition $m^{3}+n^{3}>(m+n)^{2}$. If $m+n=4$, then $(m, n)=(1,3),(2,2),(3,1)$. The pair $(m, n)=(2,2)$ doesn't satisfy the condition. The pairs $(m, n)=(1,3),(3,1)$ satisfy the condition and we can readily check that $m^{3}+n^{3} \geq(m+$ $n)^{2}+10$. If $m+n \geq 5$ then we will show that $$ m^{2}+n^{2}-m n-m-n \geq 2 $$ which is equivalent to $$ (m-n)^{2}+(m-1)^{2}+(n-1)^{2} \geq 6 $$ If at least one of the numbers $m, n$ is greater or equal to 4 then $(m-1)^{2} \geq 9$ or $(n-1)^{2} \geq 9$ hence the desired result holds. As a result, it remains to check what happens if $m \leq 3$ and $n \leq 3$. Using the condition $m+n \geq 5$ we have that all such pairs are $(m, n)=(2,3),(3,2),(3,3)$. All of them satisfy the condition and also the inequality $m^{2}+n^{2}-m n-m-n \geq 2$, thus we have the desired result.

10

Inequalities

math-word-problem

Yes

Yes

olympiads

A 4. Let $k>1, n>2018$ be positive integers, and let $n$ be odd. The nonzero rational numbers $x_{1}$, $x_{2}, \ldots, x_{n}$ are not all equal and satisfy $$ x_{1}+\frac{k}{x_{2}}=x_{2}+\frac{k}{x_{3}}=x_{3}+\frac{k}{x_{4}}=\cdots=x_{n-1}+\frac{k}{x_{n}}=x_{n}+\frac{k}{x_{1}} $$ Find: a) the product $x_{1} x_{2} \ldots x_{n}$ as a function of $k$ and $n$ b) the least value of $k$, such that there exist $n, x_{1}, x_{2}, \ldots, x_{n}$ satisfying the given conditions.

Solution. a) If $x_{i}=x_{i+1}$ for some $i$ (assuming $x_{n+1}=x_{1}$ ), then by the given identity all $x_{i}$ will be equal, a contradiction. Thus $x_{1} \neq x_{2}$ and $$ x_{1}-x_{2}=k \frac{x_{2}-x_{3}}{x_{2} x_{3}} $$ Analogously $$ x_{1}-x_{2}=k \frac{x_{2}-x_{3}}{x_{2} x_{3}}=k^{2} \frac{x_{3}-x_{4}}{\left(x_{2} x_{3}\right)\left(x_{3} x_{4}\right)}=\cdots=k^{n} \frac{x_{1}-x_{2}}{\left(x_{2} x_{3}\right)\left(x_{3} x_{4}\right) \ldots\left(x_{1} x_{2}\right)} $$ Since $x_{1} \neq x_{2}$ we get $$ x_{1} x_{2} \ldots x_{n}= \pm \sqrt{k^{n}}= \pm k^{\frac{n-1}{2}} \sqrt{k} $$ If one among these two values, positive or negative, is obtained, then the other one will be also obtained by changing the sign of all $x_{i}$ since $n$ is odd. b) From the above result, as $n$ is odd, we conclude that $k$ is a perfect square, so $k \geq 4$. For $k=4$ let $n=2019$ and $x_{3 j}=4, x_{3 j-1}=1, x_{3 j-2}=-2$ for $j=1,2, \ldots, 673$. So the required least value is $k=4$. Comment by PSC. There are many ways to construct the example when $k=4$ and $n=2019$. Since $3 \mid 2019$, the idea is to find three numbers $x_{1}, x_{2}, x_{3}$ satisfying the given equations, not all equal, and repeat them as values for the rest of the $x_{i}$ 's. So, we want to find $x_{1}, x_{2}, x_{3}$ such that $$ x_{1}+\frac{4}{x_{2}}=x_{2}+\frac{4}{x_{3}}=x_{3}+\frac{4}{x_{1}} $$ As above, $x_{1} x_{2} x_{3}= \pm 8$. Suppose without loss of generality that $x_{1} x_{2} x_{3}=-8$. Then, solving the above system we see that if $x_{1} \neq 2$, then $$ x_{2}=-\frac{4}{x_{1}-2} \text { and } x_{3}=2-\frac{4}{x_{1}} $$ leading to infinitely many solutions. The example in the official solution is obtained by choosing $x_{1}=-2$. Comment by PSC. An alternative formulation of the problem's statement could be the following: Let $k>1$ be a positive integer. Suppose that there exists an odd positive integer $n>2018$ and nonzero rational numbers $x_{1}, x_{2}, \ldots, x_{n}$, not all of them equal, that satisfy $$ x_{1}+\frac{k}{x_{2}}=x_{2}+\frac{k}{x_{3}}=x_{3}+\frac{k}{x_{4}}=\cdots=x_{n-1}+\frac{k}{x_{n}}=x_{n}+\frac{k}{x_{1}} $$ Find the minimum value of $k$.

4

Algebra

math-word-problem

Yes

Yes

olympiads

A 5. Let $a, b, c, d$ and $x, y, z, t$ be real numbers such that $$ 0 \leq a, b, c, d \leq 1, \quad x, y, z, t \geq 1 \text { and } a+b+c+d+x+y+z+t=8 $$ Prove that $$ a^{2}+b^{2}+c^{2}+d^{2}+x^{2}+y^{2}+z^{2}+t^{2} \leq 28 $$ When does the equality hold?

Solution. We observe that if $u \leq v$ then by replacing $(u, v)$ with $(u-\varepsilon, v+\varepsilon)$, where $\varepsilon>0$, the sum of squares increases. Indeed, $$ (u-\varepsilon)^{2}+(v+\varepsilon)^{2}-u^{2}-v^{2}=2 \varepsilon(v-u)+2 \varepsilon^{2}>0 $$ Then, denoting $$ E(a, b, c, d, x, y, z, t)=a^{2}+b^{2}+c^{2}+d^{2}+x^{2}+y^{2}+z^{2}+t^{2} $$ and assuming without loss of generality that $a \leq b \leq c \leq d$ and $x \leq y \leq z \leq t$, we have $$ \begin{aligned} E(a, b, c, d, x, y, z, t) & \leq E(0,0,0,0, a+x, b+y, c+z, d+t) \\ & \leq E(0,0,0,0,1, b+y, c+z, a+d+x+t-1) \\ & \leq E(0,0,0,0,1,1, c+z, a+b+d+x+y+t-2) \\ & \leq E(0,0,0,0,1,1,1,5)=28 \end{aligned} $$ Note that if $(a, b, c, d, x, y, z, t) \neq(0,0,0,0,1,1,1,5)$, at least one of the above inequalities, obtained by the $\epsilon$ replacement mentioned above, should be a strict inequality. Thus, the maximum value of $E$ is 28 , and it is obtained only for $(a, b, c, d, x, y, z, t)=(0,0,0,0,1,1,1,5)$ and permutations of $a, b, c, d$ and of $x, y, z, t$.

28

Inequalities

proof

Yes

Yes

olympiads

A 7. Let $A$ be a set of positive integers with the following properties: (a) If $n$ is an element of $A$ then $n \leqslant 2018$. (b) If $S$ is a subset of $A$ with $|S|=3$ then there are two elements $n, m$ of $S$ with $|n-m| \geqslant \sqrt{n}+\sqrt{m}$. What is the maximum number of elements that $A$ can have?

Solution. Assuming $n>m$ we have $$ \begin{aligned} |n-m| \geqslant \sqrt{n}+\sqrt{m} & \Leftrightarrow(\sqrt{n}-\sqrt{m})(\sqrt{n}+\sqrt{m}) \geqslant \sqrt{n}+\sqrt{m} \\ & \Leftrightarrow \sqrt{n} \geqslant \sqrt{m}+1 . \end{aligned} $$ Let $A_{k}=\left\{k^{2}, k^{2}+1, \ldots,(k+1)^{2}-1\right\}$. Note that each $A_{k}$ can contain at most two elements of since if $n, m \in$ with $n>m$ then $$ \sqrt{n}-\sqrt{m} \leqslant \sqrt{(k+1)^{2}-1}-\sqrt{k^{2}}<(k+1)-k=1 $$ In particular, since $\subseteq A_{1} \cup \cdots \cup A_{44}$, we have $|S| \leqslant 2 \cdot 44=88$. On the other hand we claim that $A=\left\{m^{2}: 1 \leqslant m \leqslant 44\right\} \cup\left\{m^{2}+m: 1 \leqslant m \leqslant 44\right\}$ satisfies the properties and has $|A|=88$. We check property (b) as everything else is trivial. So let $r, s, t$ be three elements of $A$ and assume $r<s<t$. There are two cases for $r$. (i) If we have that $r=m^{2}$, then $t \geqslant(m+1)^{2}$ and so $\sqrt{t}-\sqrt{r} \geq 1$ verifying (b). (ii) If we have that $r=m^{2}+m$, then $t \geqslant(m+1)^{2}+(m+1)$ and $$ \begin{aligned} \sqrt{t} \geqslant \sqrt{r}+1 & \Leftrightarrow \sqrt{(m+1)^{2}+(m+1)} \geqslant \sqrt{m^{2}+m}+1 \\ & \Leftrightarrow m^{2}+3 m+2 \geqslant m^{2}+m+1+2 \sqrt{m^{2}+m} \\ & \Leftrightarrow 2 m+1 \geqslant 2 \sqrt{m^{2}+m} \\ & \Leftrightarrow 4 m^{2}+4 m+1 \geqslant 4 m^{2}+4 m . \end{aligned} $$ So property (b) holds in this case as well. ## COMBINATORICS

88

Combinatorics

math-word-problem

Yes

Yes

olympiads

C 2. A set $T$ of $n$ three-digit numbers has the following five properties: (1) No number contains the digit 0 . (2) The sum of the digits of each number is 9 . (3) The units digits of any two numbers are different. (4) The tens digits of any two numbers are different. (5) The hundreds digits of any two numbers are different. Find the largest possible value of $n$.

Solution. Let $S$ denote the set of three-digit numbers that have digit sum equal to 9 and no digit equal to 0 . We will first find the cardinality of $S$. We start from the number 111 and each element of $S$ can be obtained from 111 by a string of $6 A$ 's (which means that we add 1 to the current digit) and 2 G's (which means go to the next digit). Then for example 324 can be obtained from 111 by the string AAGAGAAA. There are in total $$ \frac{8!}{6!\cdot 2!}=28 $$ such words, so $S$ contains 28 numbers. Now, from the conditions (3), (4), (5), if $\overline{a b c}$ is in $T$ then each of the other numbers of the form $\overline{* c}$ cannot be in $T$, neither $\overline{* * *}$ can be, nor $\overline{a * *}$. Since there are $a+b-2$ numbers of the first category, $a+c-2$ from the second and $b+c-2$ from the third one. In these three categories there are $$ (a+b-2)+(b+c-2)+(c+a-2)=2(a+b+c)-6=2 \cdot 9-6=12 $$ distinct numbers that cannot be in $T$ if $\overline{a b c}$ is in $T$. So, if $T$ has $n$ numbers, then $12 n$ are the forbidden ones that are in $S$, but each number from $S$ can be a forbidden number no more than three times, once for each of its digits, so $$ n+\frac{12 n}{3} \leq 28 \Longleftrightarrow n \leq \frac{28}{5} $$ and since $n$ is an integer, we get $n \leq 5$. A possible example for $n=5$ is $$ T=\{144,252,315,423,531\} $$ Comment by PSC. It is classical to compute the cardinality of $S$ and this can be done in many ways. In general, the number of solutions of the equation $$ x_{1}+x_{2}+\cdots+x_{k}=n $$ in positive integers, where the order of $x_{i}$ matters, is well known that equals to $\binom{n-1}{k-1}$. In our case, we want to count the number of positive solutions to we want to count the number of positive solutions to $a+b+c=9$. By the above, this equals to $\binom{9-1}{3-1}=28$. Using the general result above, we can also find that there are $a+b-2$ numbers of the form $\overline{* * c}$.

5

Combinatorics

math-word-problem

Yes

Yes

olympiads

C2 Five players $(A, B, C, D, E)$ take part in a bridge tournament. Every two players must play (as partners) against every other two players. Any two given players can be partners not more than once per day. What is the least number of days needed for this tournament?

Solution: A given pair must play with three other pairs and these plays must be in different days, so at three days are needed. Suppose that three days suffice. Let the pair $A B$ play against $C D$ on day $x$. Then $A B-D E$ and $C D-B E$ cannot play on day $x$. Then one of the other two plays of $D E$ (with $A C$ and $B C$ ) must be on day $x$. Similarly, one of the plays of $B E$ with $A C$ or $A D$ must be on day $x$. Thus, two of the plays in the chain $B C-D E-A C-B E-A D$ are on day $x$ (more than two among these cannot be on one day). Consider the chain $A B-C D-E A-B D-C E-A B$. At least three days are needed for playing all the matches within it. For each of these days we conclude (as above) that there are exactly two of the plays in the chain $B C-D E-A C-B E-A D-B C$ on that day. This is impossible, as this chain consists of five plays. It remains to show that four days will suffice: Day 1: $A B-C D, A C-D E, A D-C E, A E-B C$ Day 2: $A B-D E, A C-B D, A D-B C, B E-C D$ Day 3: $A B-C E, A D-B E, A E-B D, B C-D E$ Day 4: $A C-B E, A E-C D, B D-C E$. Remark: It is possible to have 5 games in one day (but not on each day).

4

Combinatorics

math-word-problem

Yes

Yes

olympiads

C3 a) In how many ways can we read the word SARAJEVO from the table below, if it is allowed to jump from cell to an adjacent cell (by vertex or a side) cell?  b) After the letter in one cell was deleted, only 525 ways to read the word SARAJEVO remained. Find all possible positions of that cell.

Solution: In the first of the tables below the number in each cell shows the number of ways to reach that cell from the start (which is the sum of the quantities in the cells, from which we can come), and in the second one are the number of ways to arrive from that cell to the end (which is the sum of the quantities in the cells, to which we can go). a) The answer is 750 , as seen from the second table. b) If we delete the letter in a cell, the number of ways to read SARAJEVO will decrease by the product of the numbers in the corresponding cell in the two tables. As $750-525=225$, this product has to be 225. This happens only for two cells on the third row. Here is the table with the products:

750

Combinatorics

math-word-problem

Yes

Yes

olympiads

G2 In a right trapezoid $A B C D(A B \| C D)$ the angle at vertex $B$ measures $75^{\circ}$. Point $H$ is the foot of the perpendicular from point $A$ to the line $B C$. If $B H=D C$ and $A D+A H=8$, find the area of $A B C D$.

Solution: Produce the legs of the trapezoid until they intersect at point $E$. The triangles $A B H$ and $E C D$ are congruent (ASA). The area of $A B C D$ is equal to area of triangle $E A H$ of hypotenuse $$ A E=A D+D E=A D+A H=8 $$ Let $M$ be the midpoint of $A E$. Then $$ M E=M A=M H=4 $$ and $\angle A M H=30^{\circ}$. Now, the altitude from $H$ to AM equals one half of $M H$, namely 2. Finally, the area is 8 . 

8

Geometry

math-word-problem

Yes

Yes

olympiads

NT2 A group of $n>1$ pirates of different age owned total of 2009 coins. Initially each pirate (except for the youngest one) had one coin more than the next younger. a) Find all possible values of $n$. b) Every day a pirate was chosen. The chosen pirate gave a coin to each of the other pirates. If $n=7$, find the largest possible number of coins a pirate can have after several days.

## Solution: a) If $n$ is odd, then it is a divisor of $2009=7 \times 7 \times 41$. If $n>49$, then $n$ is at least $7 \times 41$, while the average pirate has 7 coins, so the initial division is impossible. So, we can have $n=7, n=41$ or $n=49$. Each of these cases is possible (e.g. if $n=49$, the average pirate has 41 coins, so the initial amounts are from $41-24=17$ to $41+24=65$ ). If $n$ is even, then 2009 is multiple of the sum $S$ of the oldest and the youngest pirate. If $S<7 \times 41$, then $S$ is at most 39 and the pairs of pirates of sum $S$ is at least 41 , so we must have at least 82 pirates, a contradiction. So we can have just $S=7 \times 41=287$ and $S=49 \times 41=2009$; respectively, $n=2 \times 7=14$ or $n=2 \times 1=2$. Each of these cases is possible (e.g. if $n=14$, the initial amounts are from $144-7=137$ to $143+7=150$ ). In total, $n$ is one of the numbers $2,7,13,41$ and 49 . b) If $n=7$, the average pirate has $7 \times 41=287$ coins, so the initial amounts are from 284 to 290; they have different residues modulo 7. The operation decreases one of the amounts by 6 and increases the other ones by 1 , so the residues will be different at all times. The largest possible amount in one pirate's possession will be achieved if all the others have as little as possible, namely $0,1,2,3,4$ and 5 coins (the residues modulo 7 have to be different). If this happens, the wealthiest pirate will have $2009-14=1994$ coins. Indeed, this can be achieved e.g. if every day (until that moment) the coins are given by the second wealthiest: while he has more than 5 coins, he can provide the 6 coins needed, and when he has no more than five, the coins at the poorest six pirates have to be $0,1,2,3,4,5$. Thus, $n=1994$ can be achieved.

1994

Number Theory

math-word-problem

Yes

Yes

olympiads

C2 Can we divide an equilateral triangle $\triangle A B C$ into 2011 small triangles using 122 straight lines? (there should be 2011 triangles that are not themselves divided into smaller parts and there should be no polygons which are not triangles)

## Solution Firstly, for each side of the triangle, we draw 37 equidistant, parallel lines to it. In this way we get $38^{2}=1444$ triangles. Then we erase 11 lines which are closest to the vertex $A$ and parallel to the side $B C$ and we draw 21 lines perpendicular to $B C$, the first starting from the vertex $A$ and 10 on each of the two sides, the lines which are closest to the vertex $A$, distributed symmetrically. In this way we get $26 \cdot 21+10=$ 556 new triangles. Therefore we obtain a total of 2000 triangles and we have used $37 \cdot 3-11+21=121$ lines. Let $D$ be the $12^{\text {th }}$ point on side $A B$, starting from $B$ (including it). The perpendicular to $B C$ passing through $D$ will be the last line we draw. In this way we obtain the required configuration.

2011

Geometry

math-word-problem

Yes

Yes

olympiads

C4 In a group of $n$ people, each one had a different ball. They performed a sequence of swaps; in each swap, two people swapped the ball they had at that moment. Each pair of people performed at least one swap. In the end each person had the ball he/she had at the start. Find the least possible number of swaps, if: $a$ ) $n=5$; b) $n=6$.

## Solution We will denote the people by $A, B, C, \ldots$ and their initial balls by the corresponding small letters. Thus the initial state is $A a, B b, C c, D d, E e(, F f)$. A swap is denoted by the (capital) letters of the people involved. a) Five people form 10 pairs, so at least 10 swaps are necessary. In fact, 10 swaps are sufficient: Swap $A B$, then $B C$, then $C A$; the state is now $A a, B c, C b, D d, E e$. Swap $A D$, then $D E$, then $E A$; the state is now $A a, B c, C b, D e, E d$. Swap $B E$, then $C D$; the state is now $A a, B d, C e, D b, E c$. Swap $B D$, then $C E$; the state is now $A a, B b, C c, D d, E e$. All requirements are fulfilled now, so the answer is 10 . b) Six people form 15 pairs, so at least 15 swaps are necessary. We will prove that the final number of swaps must be even. Call a pair formed by a ball and a person inverted if letter of the ball lies after letter of the person in the alphabet. Let $T$ be the number of inverted pairs; at the start we have $T=0$. Each swap changes $T$ by 1 , so it changes the parity of $T$. Since in the end $T=0$, the total number of swaps must be even. Hence, at least 16 swaps are necessary. In fact 16 swaps are sufficient: Swap $A B$, then $B C$, then $C A$; the state is now $A a, B c, C b, D d, E e, F f$. Swap $A D$, then $D E$, then $E A$; the state is now $A a, B c, C b, D e, E d, F f$. Swap $F B$, then $B E$, then $E F$; the state is now $A a, B d, C b, D e, E c, F f$. Swap $F C$, then $C D$, then $D F$; the state is now $A a, B d, C e, D b, E c, F f$. Swap $B D$, then $C E$, then twice $A F$, the state is now $A a, B b, C c, D d, E e, F f$. All requirements are fulfilled now, so the answer is 16 .

16

Combinatorics

math-word-problem

Yes

Yes

olympiads

C5 A set $S$ of natural numbers is called good, if for each element $x \in S, x$ does not divide the sum of the remaining numbers in $S$. Find the maximal possible number of elements of a good set which is a subset of the set $A=\{1,2,3, \ldots, 63\}$.

Solution Let set $B$ be the good subset of $A$ which have the maximum number of elements. We can easily see that the number 1 does not belong to $B$ since 1 divides all natural numbers. Based on the property of divisibility, we know that $x$ divides the sum of the remaining numbers if and only if $x$ divides the sum of all numbers in the set $B$. If $B$ has exactly 62 elements, than $B=\{2,3,4, \ldots, 62\}$, but this set can't be good since the sum of its elements is 2015 which is divisible by 5 . Therefore $B$ has at most 61 elements. Now we are looking for the set, whose elements does not divide their sum, so the best way to do that is making a sum of elements be a prime number. $2+3+4+\ldots+63=2015$ and if we remove the number 4, we will obtain the prime number 2011. Hence the set $B=\{2,3,5,6,7, \ldots, 63\}$ is a good one. We conclude that our number is 61 .

61

Number Theory

math-word-problem

Yes

Yes

olympiads

G3. Let $A B C D E F$ be a regular hexagon. The points $\mathrm{M}$ and $\mathrm{N}$ are internal points of the sides $\mathrm{DE}$ and $\mathrm{DC}$ respectively, such that $\angle A M N=90^{\circ}$ and $A N=\sqrt{2} \cdot C M$. Find the measure of the angle $\angle B A M$.

## Solution Since $A C \perp C D$ and $A M \perp M N$ the quadrilateral $A M N C$ is inscribed. So, we have $$ \angle M A N=\angle M C N $$ Let $P$ be the projection of the point $M$ on the line $C D$. The triangles $A M N$ and $C P M$ are similar implying $$ \frac{A M}{C P}=\frac{M N}{P M}=\frac{A N}{C M}=\sqrt{2} $$ So, we have $$ \frac{M P}{M N}=\frac{1}{\sqrt{2}} \Rightarrow \angle M N P=45^{\circ} $$  Figure 4 Hence we have $$ \angle C A M=\angle M N P=45^{\circ} $$ and finally, we obtain $$ \angle B A M=\angle B A C+\angle C A M=75^{\circ} $$

75

Geometry

math-word-problem

Yes

Yes

olympiads

G6. A point $O$ and the circles $k_{1}$ with center $O$ and radius $3, k_{2}$ with center $O$ and radius 5, are given. Let $A$ be a point on $k_{1}$ and $B$ be a point on $k_{2}$. If $A B C$ is equilateral triangle, find the maximum value of the distance $O C$.

## Solution It is easy to see that the points $O$ and $C$ must be in different semi-planes with respect to the line $A B$. Let $O P B$ be an equilateral triangle ( $P$ and $C$ on the same side of $O B$ ). Since $\angle P B C$ $=60^{\circ}-\angle A B P$ and $\angle O B A=60^{\circ}-\angle A B P$, then $\angle P B C=\angle O B A$. Hence the triangles $A O B$ and $C P B$ are equal and $P C=O A$. From the triangle $O P C$ we have $$ O C \leq O P+P C=O B+O A=8 $$ Hence, the maximum yalue of the distance $O C$ is 8 (when the point $P$ lies on $O C$ )  Figure 8  Figure 9 ## Complement to the solution Indeed there exists a triangle $\mathrm{OAB}$ with $\mathrm{OA}=3, \mathrm{OB}=5$ and $\mathrm{OC}=8$. To construct such a triangle, let's first consider a point $M$ on the minor arc $\widehat{A_{0} B_{0}}$ of the circumference $\left(c_{0}\right)$ of an arbitrary equilateral triangle $A_{0} B_{0} C_{0}$. As $\mathrm{M}$ moves along $\widehat{\mathrm{A}_{0} \mathrm{~B}_{0}}$ from the midpoint position $\mathrm{M}_{0}$ towards $\mathrm{A}_{0}$, the ratio $\frac{\mathrm{MA}_{0}}{\mathrm{MB}_{0}}$ takes on all the decreasing values from 1 to 0 . Thus there exists a position of $\mathrm{M}$ such that $\frac{\mathrm{MA}_{0}}{\mathrm{MB}_{0}}=\frac{3}{5}$. Now a homothesy centered at the center of $\left(\mathrm{c}_{0}\right)$ can take $\mathrm{A}_{0}, \mathrm{~B}_{0}, \mathrm{C}_{0}, \mathrm{M}$ to the new positions $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{O}$ so that $\mathrm{OA}=3$ and $\mathrm{OB}=5$. Then, since $\mathrm{C}$ lies on the minor arc $\overparen{\mathrm{AB}}$ of the circumference (c) of the equilateral triangle $\mathrm{ABC}$ we get $\mathrm{OC}=\mathrm{OA}+\mathrm{OB}=3+5=8$ as wanted, (figure 9).

8

Geometry

math-word-problem

Yes

Yes

olympiads

NT2. Find all natural numbers $n$ such that $5^{n}+12^{n}$ is perfect square.

## Solution By checking the cases $n=1,2,3$ we get the solution $n=2$ and $13^{2}=5^{2}+12^{2}$. If $n=2 k+1$ is odd, we consider the equation modulo 5 and we obtain $$ \begin{aligned} x^{2} & \equiv 5^{2 k+1}+12^{2 k+1}(\bmod 5) \equiv 2^{2 k} \cdot 2(\bmod 5) \\ & \equiv(-1)^{k} \cdot 2(\bmod 5) \equiv \pm 2(\bmod 5) \end{aligned} $$ This is not possible, because the square residue of any natural number module 5 is 0,1 or 4. Therefore $n$ is even and $x^{2}=5^{2 k}+12^{2 k}$. Rearrange this equation in the form $$ 5^{2 k}=\left(x-12^{k}\right)\left(x+12^{k}\right) $$ If 5 divides both factors on the right, it must also divide their difference, that is $$ 5 \mid\left(x+12^{k}\right)-\left(x-12^{k}\right)=2 \cdot 12^{k} $$ which is not possible. Therefore we must have $$ x-12^{k}=1 \text { and } x+12^{k}=5^{2 k} $$ By adding the above equalities we get $$ 5^{2 k}-1=2 \cdot 12^{k} $$ For $k \geq 2$, we have the inequality $$ 25^{k}-1>24^{k}=2^{k} \cdot 12^{k}>2 \cdot 12^{k} $$ Thus we conclude that there exists a unique solution to our problem, namely $n=2$.

2

Number Theory

math-word-problem

Yes

Yes

olympiads

NT4. Find all the three digit numbers $\overline{a b c}$ such that $$ \overline{a b c}=a b c(a+b+c) $$

## Solution We will show that the only solutions are 135 and 144 . We have $a>0, b>0, c>0$ and $$ 9(11 a+b)=(a+b+c)(a b c-1) $$ - If $a+b+c \equiv 0(\bmod 3)$ and $a b c-1 \equiv 0(\bmod 3)$, then $a \equiv b \equiv c \equiv 1(\bmod 3)$ and $11 a+b \equiv 0(\bmod 3)$. It follows now that $$ a+b+c \equiv 0(\bmod 9) ; \text { or } a b c-1 \equiv 0(\bmod 9) $$ - If . $a b c-1 \equiv 0(\bmod 9)$ we have $11 a+b=(a+b+c) k$, where $k$ is an integer and is easy to see that we must have $19$. Now we will deal with the case when $a+b+c \equiv 0(\bmod 9)$ or $a+b+c=9 l$, where $l$ is an integer. - If $l \geq 2$ we have $a+b+c \geq 18, \max \{a, b, c\} \geq 6$ and it is easy to see that $a b c \geq 72$ and $a b c(a+b+c)>1000$,so the case $l \geq 2$ is impossible. - If $l=1$ we have $$ 11 a+b=a b c-1 \text { or } 11 a+b+1=a b c \leq\left(\frac{a+b+c}{3}\right)^{3}=27 $$ So we have only two cases: $a=1$ or $a=2$. - If $a=1$, we have $b+c=8$ and $11+b=b c-1$ or $b+(c-1)=7$ and $b(c-1)=12$ and the solutions are $(a, b, c)=(1,3,5)$ and $(a, b, c)=(1,4,4)$, and the answer is 135 and 144. - If $a=2$ we have $b(2 c-1)=23$ and there is no solution for the problem.

135144

Number Theory

math-word-problem

Yes

Yes

olympiads

A4 Let $x, y$ be positive real numbers such that $x^{3}+y^{3} \leq x^{2}+y^{2}$. Find the greatest possible value of the product $x y$.

## Solution 1 We have $(x+y)\left(x^{2}+y^{2}\right) \geq(x+y)\left(x^{3}+y^{3}\right) \geq\left(x^{2}+y^{2}\right)^{2}$, hence $x+y \geq x^{2}+y^{2}$. Now $2(x+y) \geq(1+1)\left(x^{2}+y^{2}\right) \geq(x+y)^{2}$, thus $2 \geq x+y$. Because $x+y \geq 2 \sqrt{x y}$, we will obtain $1 \geq x y$. Equality holds when $x=y=1$. So the greatest possible value of the product $x y$ is 1 .

1

Inequalities

math-word-problem

Yes

Yes

olympiads

A6 Let $x_{i}>1$, for all $i \in\{1,2,3, \ldots, 2011\}$. Prove the inequality $\sum_{i=1}^{2011} \frac{x_{i}^{2}}{x_{i+1}-1} \geq 8044$ where $x_{2012}=x_{1}$. When does equality hold?

## Solution 1 Realize that $\left(x_{i}-2\right)^{2} \geq 0 \Leftrightarrow x_{i}^{2} \geq 4\left(x_{i}-1\right)$. So we get: $\frac{x_{1}^{2}}{x_{2}-1}+\frac{x_{2}^{2}}{x_{3}-1}+\ldots+\frac{x_{2011}^{2}}{x_{1}-1} \geq 4\left(\frac{x_{1}-1}{x_{2}-1}+\frac{x_{2}-1}{x_{3}-1}+\ldots+\frac{x_{2011}-1}{x_{1}-1}\right)$. By $A M-G M$ : $\frac{x_{1}-1}{x_{2}-1}+\frac{x_{2}-1}{x_{3}-1}+\ldots+\frac{x_{2011}-1}{x_{1}-1} \geq 2011 \cdot \sqrt[2011]{\frac{x_{1}-1}{x_{2}-1} \cdot \frac{x_{2}-1}{x_{3}-1} \cdot \ldots \cdot \frac{x_{2011}-1}{x_{1}-1}}=2011$ Finally, we obtain that $\frac{x_{1}^{2}}{x_{2}-1}+\frac{x_{2}^{2}}{x_{3}-1}+\ldots+\frac{x_{2011}^{2}}{x_{1}-1} \geq 8044$. Equality holds when $\left(x_{i}-2\right)^{2}=0,(\forall) i=\overline{1,2011}$, or $x_{1}=x_{2}=\ldots=x_{2011}=2$.

8044

Inequalities

proof

Yes

Yes

olympiads

A7 Let $a, b, c$ be positive real numbers with $a b c=1$. Prove the inequality: $$ \frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1}+\frac{2 b^{2}+\frac{1}{b}}{c+\frac{1}{b}+1}+\frac{2 c^{2}+\frac{1}{c}}{a+\frac{1}{c}+1} \geq 3 $$

## Solution 1 By $A M-G M$ we have $2 x^{2}+\frac{1}{x}=x^{2}+x^{2}+\frac{1}{x} \geq 3 \sqrt[3]{\frac{x^{4}}{x}}=3 x$ for all $x>0$, so we have: $\sum_{\text {cyc }} \frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1} \geq \sum_{c y c} \frac{3 a}{1+b+b c}=3\left(\sum_{c y c} \frac{a^{2}}{1+a+a b}\right) \geq \frac{3(a+b+c)^{2}}{3+a+b+c+a b+b c+c a}$. By $A M-G M$ we have $a b+b c+c a \geq 3$ and $a+b+c \geq 3$. But $3\left(a^{2}+b^{2}+c^{2}\right) \geq(a+b+c)^{2} \geq$ $3(a+b+c)$. So $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a \geq 3+a+b+c+a b+b c+c a$. Hence $\sum_{c y c} \frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1} \geq \frac{3(a+b+c)^{2}}{3+a+b+c+a b+b c+c a} \geq \frac{3(a+b+c)^{2}}{(a+b+c)^{2}}=3$.

3

Inequalities

proof

Yes

Yes

olympiads

A9 Consider an integer $n \geq 4$ and a sequence of real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$. An operation consists in eliminating all numbers not having the rank of the form $4 k+3$, thus leaving only the numbers $x_{3}, x_{7}, x_{11}, \ldots$ (for example, the sequence $4,5,9,3,6,6,1,8$ produces the sequence 9,1 . Upon the sequence $1,2,3, \ldots, 1024$ the operation is performed successively for 5 times. Show that at the end only 1 number remains and find this number.

## Solution After the first operation 256 number remain; after the second one, 64 are left, then 16, next 4 and ultimately only one number. Notice that the 256 numbers left after the first operation are $3,7, \ldots, 1023$, hence they are in arithmetical progression of common difference 4. Successively, the 64 numbers left after the second operation are in arithmetical progression of ratio 16 and so on. Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be the first term in the 5 sequences obtained after each of the 5 operations. Thus $a_{1}=3$ and $a_{5}$ is the requested number. The sequence before the fifth operation has 4 numbers, namely $$ a_{4}, a_{4}+256, a_{4}+512, a_{4}+768 $$ and $a_{5}=a_{4}+512$. Similarly, $a_{4}=a_{3}+128, a_{3}=a_{2}+32, a_{2}=a_{1}+8$. Summing up yields $a_{5}=a_{1}+8+32+128+512=3+680=683$. ### 2.2 Combinatorics

683

Combinatorics

math-word-problem

Yes

Yes

olympiads

G3 The vertices $A$ and $B$ of an equilateral $\triangle A B C$ lie on a circle $k$ of radius 1 , and the vertex $C$ is inside $k$. The point $D \neq B$ lies on $k, A D=A B$ and the line $D C$ intersects $k$ for the second time in point $E$. Find the length of the segment $C E$.

## Solution As $A D=A C, \triangle C D A$ is isosceles. If $\varangle A D C=\varangle A C D=\alpha$ and $\varangle B C E=\beta$, then $\beta=120^{\circ}-\alpha$. The quadrilateral $A B E D$ is cyclic, so $\varangle A B E=180^{\circ}-\alpha$. Then $\varangle C B E=$ $120^{\circ}-\alpha$ so $\varangle C B E=\beta$. Thus $\triangle C B E$ is isosceles, so $A E$ is the perpendicular bisector of $B C$, so it bisects $\varangle B A C$. Now the arc $B E$ is intercepted by a $30^{\circ}$ inscribed angle, so it measures $60^{\circ}$. Then $B E$ equals the radius of $k$, namely 1 . Hence $C E=B E=1$. 

1

Geometry

math-word-problem

Yes

Yes

olympiads

NT3 Let $s(a)$ denote the sum of digits of a given positive integer $a$. The sequence $a_{1}, a_{2}, \ldots a_{n}, \ldots$ of positive integers is such that $a_{n+1}=a_{n}+s\left(a_{n}\right)$ for each positive integer $n$. Find the greatest possible $n$ for which it is possible to have $a_{n}=2008$.

## Solution Since $a_{n-1} \equiv s\left(a_{n-1}\right)$ (all congruences are modulo 9 ), we have $2 a_{n-1} \equiv a_{n} \equiv 2008 \equiv 10$, so $a_{n-1} \equiv 5$. But $a_{n-1}<2008$, so $s\left(a_{n-1}\right) \leq 28$ and thus $s\left(a_{n-1}\right)$ can equal 5,14 or 23 . We check $s(2008-5)=s(2003)=5, s(2008-14)=s(1994)=23, s(2008-23)=s(1985)=$ 23. Thus $a_{n-1}$ can equal 1985 or 2003 . As above $2 a_{n-2} \equiv a_{n-1} \equiv 5 \equiv 14$, so $a_{n-2} \equiv 7$. But $a_{n-2}<2003$, so $s\left(a_{n-2}\right) \leq 28$ and thus $s\left(a_{n-2}\right)$ can equal 16 or 25 . Checking as above we see that the only possibility is $s(2003-25)=s(1978)=25$. Thus $a_{n-2}$ can be only 1978. Now $2 a_{n-3} \equiv a_{n-2} \equiv 7 \equiv 16$ and $a_{n-3} \equiv 8$. But $s\left(a_{n-3}\right) \leq 27$ and thus $s\left(a_{n-3}\right)$ can equal 17 or 26 . The check works only for $s(1978-17)=s(1961)=17$. Thus $a_{n-3}=1961$ and similarly $a_{n-4}=1939 \equiv 4, a_{n-5}=1919 \equiv 2$ (if they exist). The search for $a_{n-6}$ requires a residue of 1 . But $a_{n-6}<1919$, so $s\left(a_{n-6}\right) \leq 27$ and thus $s\left(a_{n-6}\right)$ can be equal only to 10 or 19 . The check fails for both $s(1919-10)=s(1909)=19$ and $s(1919-19)=s(1900)=10$. Thus $n \leq 6$ and the case $n=6$ is constructed above (1919, 1939, 1961, 1978, 2003, 2008).

6

Number Theory

math-word-problem

Yes

Yes

olympiads

NT4 Find all integers $n$ such that $n^{4}+8 n+11$ is a product of two or more consecutive integers.

Solution We will prove that $n^{4}+8 n+11$ is never a multiple of 3 . This is clear if $n$ is a multiple of 3 . If $n$ is not a multiple of 3 , then $n^{4}+8 n+11=\left(n^{4}-1\right)+12+8 n=(n-1)(n+1)\left(n^{2}+1\right)+12+8 n$, where $8 n$ is the only term not divisible by 3 . Thus $n^{4}+8 n+11$ is never the product of three or more integers. It remains to discuss the case when $n^{4}+8 n+11=y(y+1)$ for some integer $y$. We write this as $4\left(n^{4}+8 n+11\right)=4 y(y+1)$ or $4 n^{4}+32 n+45=(2 y+1)^{2}$. A check shows that among $n= \pm 1$ and $n=0$ only $n=1$ satisfies the requirement, as $1^{4}+8 \cdot 1+11=20=4 \cdot 5$. Now let $|n| \geq 2$. The identities $4 n^{2}+32 n+45=\left(2 n^{2}-2\right)^{2}+8(n+2)^{2}+9$ and $4 n^{4}+32 n+45=$ $\left(2 n^{2}+8\right)^{2}-32 n(n-1)-19$ indicate that for $|n| \geq 2,2 n^{2}-2<2 y+1<2 n^{2}+8$. But $2 y+1$ is odd, so it can equal $2 n^{2} \pm 1 ; 2 n^{2}+3 ; 2 n^{2}+5$ or $2 n^{2}+7$. We investigate them one by one. If $4 n^{4}+32 n+45=\left(2 n^{2}-1\right)^{2} \Rightarrow n^{2}+8 n+11=0 \Rightarrow(n+4)^{2}=5$, which is impossible, as 5 is not a perfect square. If $4 n^{4}+32 n+45=\left(2 n^{2}+1\right)^{2} \Rightarrow n^{2}-8 n-11=0 \Rightarrow(n-4)^{2}=27$ which also fails. Also $4 n^{4}+32 n+45=\left(2 n^{2}+3\right)^{2} \Rightarrow 3 n^{2}-8 n-9=0 \Rightarrow 9 n^{2}-24 n-27=0 \Rightarrow(3 n-4)^{2}=43$ fails. If $4 n^{4}+32 n+45=\left(2 n^{2}+5\right)^{2} \Rightarrow 5 n^{2}-8 n=5 \Rightarrow 25 n^{2}-40 n=25 \Rightarrow(5 n-4)^{2}=41$ which also fails. Finally, if $4 n^{4}+32 n+45=\left(2 n^{2}+7\right)^{2}$, then $28 n^{2}-32 n+4=0 \Rightarrow 4(n-1)(7 n-1)=0$, whence $n=1$ that we already found. Thus the only solution is $n=1$.

1

Number Theory

math-word-problem

Yes

Yes

olympiads

NT6 Let $f: \mathbb{N} \rightarrow \mathbb{R}$ be a function, satisfying the following condition: for every integer $n>1$, there exists a prime divisor $p$ of $n$ such that $f(n)=f\left(\frac{n}{p}\right)-f(p)$. If $$ f\left(2^{2007}\right)+f\left(3^{2008}\right)+f\left(5^{2009}\right)=2006 $$ determine the value of $$ f\left(2007^{2}\right)+f\left(2008^{3}\right)+f\left(2009^{5}\right) $$

## Solution If $n=p$ is prime number, we have $$ f(p)=f\left(\frac{p}{p}\right)-f(p)=f(1)-f(p) $$ i.e. $$ f(p)=\frac{f(1)}{2} $$ If $n=p q$, where $p$ and $q$ are prime numbers, then $$ f(n)=f\left(\frac{n}{p}\right)-f(p)=f(q)-f(p)=\frac{f(1)}{2}-\frac{f(1)}{2}=0 $$ If $n$ is a product of three prime numbers, we have $$ f(n)=f\left(\frac{n}{p}\right)-f(p)=0-f(p)=-f(p)=-\frac{f(1)}{2} $$ With mathematical induction by a number of prime multipliers we shell prove that: if $n$ is a product of $k$ prime numbers then $$ f(n)=(2-k) \frac{f(1)}{2} $$ For $k=1$, clearly the statement (2), holds. Let statement (2) holds for all integers $n$, where $n$ is a product of $k$ prime numbers. Now let $n$ be a product of $k+1$ prime numbers. Then we have $n=n_{1} p$, where $n_{1}$ is a product of $k$ prime numbers. So $$ f(n)=f\left(\frac{n}{p}\right)-f(p)=f\left(n_{1}\right)-f(p)=(2-k) \frac{f(1)}{2}-\frac{f(1)}{2}=(2-(k+1)) \frac{f(1)}{2} $$ So (2) holds for every integer $n>1$. Now from $f\left(2^{2007}\right)+f\left(3^{2008}\right)+f\left(5^{2009}\right)=2006$ and because of (2) we have $$ \begin{aligned} 2006 & =f\left(2^{2007}\right)+f\left(3^{2008}\right)+f\left(5^{2009}\right) \\ & =\frac{2-2007}{2} f(1)+\frac{2-2008}{2} f(1)+\frac{2-2009}{2} f(1)=-\frac{3 \cdot 2006}{2} f(1) \end{aligned} $$ i.e. $$ f(1)=-\frac{2}{3} $$ Since $$ 2007=3^{2} \cdot 223,2008=2^{3} \cdot 251,2009=7^{2} \cdot 41 $$ and because of (2) and (3), we get $$ \begin{aligned} f\left(2007^{2}\right)+f\left(2008^{3}\right)+f\left(2009^{5}\right) & =\frac{2-6}{2} f(1)+\frac{2-12}{2} f(1)+\frac{2-15}{2} f(1) \\ & =-\frac{27}{2} f(1)=-\frac{27}{2} \cdot\left(-\frac{2}{3}\right)=9 \end{aligned} $$

9

Number Theory

math-word-problem

Yes

Yes

olympiads

NT7 Determine the minimal prime number $p>3$ for which no natural number $n$ satisfies $$ 2^{n}+3^{n} \equiv 0(\bmod p) $$

## Solution We put $A(n)=2^{n}+3^{n}$. From Fermat's little theorem, we have $2^{p-1} \equiv 1(\bmod p)$ and $3^{p-1} \equiv 1(\bmod p)$ from which we conclude $A(n) \equiv 2(\bmod p)$. Therefore, after $p-1$ steps at most, we will have repetition of the power. It means that in order to determine the minimal prime number $p$ we seek, it is enough to determine a complete set of remainders $S(p)=\{0,1, \ldots, p-1\}$ such that $2^{n}+3^{n} \not \equiv 0(\bmod p)$, for every $n \in S(p)$. For $p=5$ and $n=1$ we have $A(1) \equiv 0(\bmod 5)$. For $p=7$ and $n=3$ we have $A(3) \equiv 0(\bmod 7)$. For $p=11$ and $n=5$ we have $A(5) \equiv 0(\bmod 11)$. For $p=13$ and $n=2$ we have $A(2) \equiv 0(\bmod 13)$. For $p=17$ and $n=8$ we have $A(8) \equiv 0(\bmod 17)$. For $p=19$ we have $A(n) \not \equiv 0(\bmod 19)$, for all $n \in S(19)$. Hence the minimal value of $p$ is 19 .

19

Number Theory

math-word-problem

Yes

Yes

olympiads

ALG 5. Let $A B C$ be a scalene triangle with $B C=a, A C=b$ and $A B=c$, where $a_{r} b, c$ are positive integers. Prove that $$ \left|a b^{2}+b c^{2}+c a^{2}-a^{2} b-b^{2} c-c^{2} a\right| \geq 2 $$

Solution. Denote $E=a b^{2}+b c^{2}+c a^{2}-a^{2} b-b^{2} c-c^{2} a$. We have $$ \begin{aligned} E= & \left(a b c-c^{2} a\right)+\left(c a^{2}-a^{2} b\right)+\left(b c^{2}-b^{2} c\right)+\left(a b^{2}-a b c\right)= \\ & (b-c)\left(a c-a^{2}-b c+a b\right)=(b-c)\left(a a^{2}-b\right)(c-a) \end{aligned} $$ So, $|E|=|a-b| \cdot|b-c| \cdot|c-a|$. By hypothesis each factor from $|E|$ is a positive integer. We shall prove that at least one factor from $|E|$ is greater than 1. Suppose that $|a-b|=|b-c|=|c-a|=1$. It follows that the numbers $a-b, b-c, c-a$ are odd. So, the number $0=(a-b)+(b-c) \div(c-a)$ is olso odd, a contradiction. Hence, $|E| \geq 1 \cdot 1 \cdot 2=2$.

2

Algebra

proof

Yes

Yes

olympiads

COM 1. In a group of 60 students: 40 speak English; 30 speak French; 8 speak all the three languages; the number of students that speak English and French but not German is equal to "the sum of the number of students that speak English and German but not French plus the number of students that speak French and German but not English; and the number of students that speak at least 2 of those fanguages is 28 . How many students speak: a) German; b) only English; c) only German?

Solution: We use the following notation. $E=\#$ students that speak English, $F=\#$ students that speak French, $G=\#$ students that speak German; $m=$ \# students that speak all the three languages, $x=\#$ students that speak English and French but not German, $y=\#$ students that speak German and French but not English, $z=\#$ students that speak English and German but not French. The conditions $x+y=z$ and $x+y+z+8=28$, imply that $z=x+y=10$, i.e. 10 students speak German and French, but not English. Then: $G+E-y-8+F-x-8-10=60$, implies that $G+70-$ $36=60$. Hence: a) $\mathrm{G}=36$; b) only English speak $40-10-8=22$ students; c) the information given is not enough to find the number of students that speak only German. This ${ }^{n}$ number can be any one from 8 to 18 . Comment: There are some mistakes in the solution. The corrections are as follows: 1. The given condition is $x=y+z($ not $x+y=z)$; thus $x=y+z=10$. 2. From $G+70-36=60$ one gets $G=26$ (not $G=36$ ). 3. One gets "only German speakers" as $G-y-z-8=8$. 4. "Only English speakers" are $E-x-z-8=22-z$, so this number can not be determined.

26

Combinatorics

math-word-problem

Yes

Incomplete

olympiads

87.3. Let $f$ be a strictly increasing function defined in the set of natural numbers satisfying the conditions $f(2)=a>2$ and $f(m n)=f(m) f(n)$ for all natural numbers $m$ and $n$. Determine the smallest possible value of $a$.

Solution. Since $f(n)=n^{2}$ is a function satisfying the conditions of the problem, the smallest posiible $a$ is at most 4. Assume $a=3$. It is easy to prove by induction that $f\left(n^{k}\right)=f(n)^{k}$ for all $k \geq 1$. So, taking into account that $f$ is strictly increasing, we get $$ \begin{gathered} f(3)^{4}=f\left(3^{4}\right)=f(81)>f(64)=f\left(2^{6}\right)=f(2)^{6} \\ =3^{6}=27^{2}>25^{2}=5^{4} \end{gathered} $$ as well as $$ \begin{aligned} & f(3)^{8}=f\left(3^{8}\right)=f(6561)<f(8192) \\ & \quad=f\left(2^{13}\right)=f(2)^{13}=3^{13}<6^{8} \end{aligned} $$ So we arrive at $5<f(3)<6$. But this is not possible, since $f(3)$ is an integer. So $a=4$.

4

Algebra

math-word-problem

Yes

Yes

olympiads

88.1. The positive integer $n$ has the following property: if the three last digits of $n$ are removed, the number $\sqrt[3]{n}$ remains. Find $n$.

Solution. If $x=\sqrt[3]{n}$, and $y, 0 \leq y1000$, and $x>31$. On the other hand, $x^{3}<1000 x+1000$, or $x\left(x^{2}-1000\right)<1000$. The left hand side of this inequality is an increasing function of $x$, and $x=33$ does not satisfy the inequality. So $x<33$. Since $x$ is an integer, $x=32$ and $n=32^{3}=32768$.

32768

Number Theory

math-word-problem

Yes

Yes

olympiads

91.1. Determine the last two digits of the number $$ 2^{5}+2^{5^{2}}+2^{5^{3}}+\cdots+2^{5^{1991}} $$ written in decimal notation.

Solution. We first show that all numbers $2^{5^{k}}$ are of the form $100 p+32$. This can be shown by induction. The case $k=1$ is clear $\left(2^{5}=32\right)$. Assume $2^{5^{k}}=100 p+32$. Then, by the binomial formula, $$ 2^{5^{k+1}}=\left(2^{5^{k}}\right)^{5}=(100 p+32)^{5}=100 q+32^{5} $$ and $$ \begin{gathered} (30+2)^{5}=30^{5}+5 \cdot 30^{4} \cdot 2+10 \cdot 30^{3} \cdot 4+10 \cdot 30^{2} \cdot 8+5 \cdot 30 \cdot 16+32 \\ =100 r+32 \end{gathered} $$ So the last two digits of the sum in the problem are the same as the last digits of the number $1991 \cdot 32$, or 12 .

12

Number Theory

math-word-problem

Yes

Yes

olympiads

92.4. Peter has many squares of equal side. Some of the squares are black, some are white. Peter wants to assemble a big square, with side equal to $n$ sides of the small squares, so that the big square has no rectangle formed by the small squares such that all the squares in the vertices of the rectangle are of equal colour. How big a square is Peter able to assemble?

Solution. We show that Peter only can make a $4 \times 4$ square. The construction is possible, if $n=4$ :  Now consider the case $n=5$. We may assume that at least 13 of the 25 squares are black. If five black squares are on one horizontal row, the remaining eight ones are distributed on the other four rows. At least one row has two black squres. A rectangle with all corners black is created. Next assume that one row has four black squares. Of the remaing 9 squares, at least three are one row. At least two of these three have to be columns having the assumed four black squares. If no row has more than four black squares, there have to be at least three rows with exactly three black squares. Denote these rows by $A, B$, and $C$. Let us call the columns in which the black squares on row $A$ lie black columns, and the other two columns white columns. If either row $B$ or row $C$ has at least two black squares which are on black columns, a rectancle with black corners arises. If both rows $B$ and $C$ have only one black square on the black columns, then both of them have two black squares on the two white columns, and they make the black corners of a rectangle. So Peter cannot make a $5 \times 5$ square in the way he wishes.

4

Combinatorics

math-word-problem

Yes

Yes

olympiads

95.2. Messages are coded using sequences consisting of zeroes and ones only. Only sequences with at most two consecutive ones or zeroes are allowed. (For instance the sequence 011001 is allowed, but 011101 is not.) Determine the number of sequences consisting of exactly 12 numbers.

Solution 1. Let $S_{n}$ be the set of acceptable sequences consisting of $2 n$ digits. We partition $S_{n}$ in subsets $A_{n}, B_{n}, C_{n}$, and $D_{n}$, on the basis of the two last digits of the sequence. Sequences ending in 00 are in $A_{n}$, those ending in 01 are in $B_{n}$, those ending in 10 are in $C_{n}$, and those ending in 11 are in $D_{n}$. Denote by $x_{n}, a_{n}, b_{n}, c_{n}$, and $d_{n}$ the number of elements in $S_{n}, A_{n}, B_{n}, C_{n}$, and $D_{n}$. We compute $x_{6}$. Because $S_{1}=\{00,01,10,11\}$, $x_{1}=4$ and $a_{1}=b_{1}=c_{1}=d_{1}=1$. Every element of $A_{n+1}$ can be obtained in a unique manner from an element of $B_{n}$ or $D_{n}$ by adjoining 00 to the end. So $a_{n+1}=b_{n}+d_{n}$. The elements of $B_{n+1}$ are similarly obtained from elements of $B_{n}, C_{n}$, and $D_{n}$ by adjoining 01 to the end. So $b_{n+1}=b_{n}+c_{n}+d_{n}$. In a similar manner we obtain the recursion formulas $c_{n+1}=a_{n}+b_{n}+c_{n}$ and $d_{n+1}=a_{n}+c_{n}$. So $a_{n+1}+d_{n+1}=\left(b_{n}+d_{n}\right)+\left(a_{n}+c_{n}\right)=x_{n}$ and $x_{n+1}=2 a_{n}+3 b_{n}+3 c_{n}+2 d_{n}=3 x_{n}-\left(a_{n}+b_{n}\right)=3 x_{n}-x_{n-1}$. Starting from the initial values $a_{1}=b_{1}=c_{1}=d_{1}=1$, we obtain $a_{2}=d_{2}=2, b_{2}=c_{2}=3$, and $x_{2}=10$. So $x_{3}=3 x_{2}-x_{1}=3 \cdot 10-4=26, x_{4}=3 \cdot 26-10=68, x_{5}=3 \cdot 68-26=178$, and $x_{6}=3 \cdot 178-68=466$.

466

Combinatorics

math-word-problem

Yes

Yes

olympiads

96.4. The real-valued function $f$ is defined for positive integers, and the positive integer a satisfies $$ \begin{gathered} f(a)=f(1995), \quad f(a+1)=f(1996), \quad f(a+2)=f(1997) \\ f(n+a)=\frac{f(n)-1}{f(n)+1} \quad \text { for all positive integers } n \end{gathered} $$ (i) Show that $f(n+4 a)=f(n)$ for all positive integers $n$. (ii) Determine the smallest possible $a$.

Solution. To prove (i), we the formula $f(n+a)=\frac{f(n)-1}{f(n)+1}$ repeatedly: $$ \begin{gathered} f(n+2 a)=f((n+a)+a)=\frac{\frac{f(n)-1}{f(n)+1}-1}{\frac{f(n)-1}{f(n)+1}+1}=-\frac{1}{f(n)} \\ f(n+4 a)=f((n+2 a)+2 a)=-\frac{1}{-\frac{1}{f(n)}}=f(n) \end{gathered} $$ (ii) If $a=1$, then $f(1)=f(a)=f(1995)=f(3+498 \cdot 4 a)=f(3)=f(1+2 a)=-\frac{1}{f(1)}$. This clearly is not possible, since $f(1)$ and $\frac{1}{f(1)}$ have equal sign. So $a \neq 1$. If $a=2$, we obtain $f(2)=f(a)=f(1995)=f(3+249 \cdot 4 a)=f(3)=f(a+1)=f(1996)=$ $f(4+249 \cdot 4 a)=f(4)=f(2+a)=\frac{f(2)-1}{f(2)+1}$, or $f(2)^{2}+f(2)=f(2)-1$. This quadratic equation in $f(2)$ has no real solutions. So $a \neq 2$. If $a=3$, we try to construct $f$ by choosing $f(1), f(2)$, and $f(3)$ arbitrarily and by computing the other values of $f$ by the recursion formula $f(n+3)=\frac{f(n)-1}{f(n)+1}$. We have to check that $f$ defined in this way satisfies the conditions of the problem. The condition $$ f(n+a)=f(n+3)=\frac{f(n)-1}{f(n)+1} $$ is valid because of the construction. Further, by (i), $$ f(n+12)=f(n+4 a)=f(n) $$ which implies $$ \begin{gathered} f(a)=f(3)=f(3+166 \cdot 12)=f(1995) \\ f(a+1)=f(4)=f(4+166 \cdot 12)=f(1996) \\ f(a+2)=f(5)=f(5+166 \cdot 12)=f(1997) \end{gathered} $$ as required. We remark that the choice $f(n)=-1$ makes $f(n+3)$ undefined, the choice $f(n)=0$ makes $f(n+3)=-1$ and $f(n+6)$ is undefined, and $f(n)=1$ makes $f(n+3)=0$ so $f(n+9)$ is undefined. In the choice of $f(1), f(2)$, and $f(3)$ we have to avoid $-1,0,1$. In conclusion, we see that $a=3$ is the smallest possible value for $a$.

3

Algebra

proof

Yes

Yes

olympiads

97.1. Let A be a set of seven positive numbers. Determine the maximal number of triples $(x, y, z)$ of elements of A satisfying $x<y$ and $x+y=z$.

Solution. Let $0<a_{1}<a_{2}<\ldots<a_{7}$ be the elements of the set $A$. If $\left(a_{i}, a_{j}, a_{k}\right)$ is a triple of the kind required in the problem, then $a_{i}<a_{j}<a_{i}+a_{j}=a_{k}$. There are at most $k-1$ pairs $\left(a_{i}, a_{j}\right)$ such that $a_{i}+a_{j}=a_{k}$. The number of pairs satisfying $a_{i}<a_{j}$ is at most $\left\lfloor\frac{k-1}{2}\right\rfloor$. The total number of pairs is at most $$ \sum_{k=3}^{7}\left\lfloor\frac{k-1}{2}\right\rfloor=1+1+2+2+3=9 $$ The value 9 can be reached, if $A=\{1,2, \ldots, 7\}$. In this case the triples $(1,2,3),(1,3,4)$, $(1,4,5),(1,5,6),(1,6,7),(2,3,5),(2,4,6),(2,5,7)$, and $(3,4,7)$ satisfy the conditions of the problem.

9

Combinatorics

math-word-problem

Yes

Yes

olympiads

99.2. Consider 7-gons inscribed in a circle such that all sides of the 7-gon are of different length. Determine the maximal number of $120^{\circ}$ angles in this kind of a 7-gon.

Solution. It is easy to give examples of heptagons $A B C D E F G$ inscribed in a circle with all sides unequal and two angles equal to $120^{\circ}$. These angles cannot lie on adjacent vertices of the heptagon. In fact, if $\angle A B C=\angle B C D=120^{\circ}$, and arc $B C$ equals $b^{\circ}$, then arcs $A B$ and $C D$ both are $120^{\circ}-b^{\circ}$ (compute angles in isosceles triangles with center of the circle as the to vertex), and $A B=C D$, contrary to the assumption. So if the heptagon has three angles of $120^{\circ}$, their vertices are, say $A, C$, and $E$. Then each of the arcs $G A B, B C D$, $D E F$ are $360^{\circ}-240^{\circ}=120^{\circ}$. The arcs are disjoint, so they cover the whole circumference. The $F$ has to coincide with $G$, and the heptagon degenerates to a hexagon. There can be at most two $120^{\circ}$ angles.

2

Geometry

math-word-problem

Yes

Yes

olympiads

00.1. In how many ways can the number 2000 be written as a sum of three positive, not necessarily different integers? (Sums like $1+2+3$ and $3+1+2$ etc. are the same.)

Solution. Since 3 is not a factor of 2000 , there has to be at least two different numbers among any three summing up to 2000 . Denote by $x$ the number of such sums with three different summands and by $y$ the number of sums with two different summands. Consider 3999 boxes consequtively numbered fron 1 to 3999 such that all boxes labelled by an odd number contain a red ball. Every way to put two blue balls in the even-numbered boxes produces a partition of 2000 in three summands. There are $\binom{1999}{2}=999 \cdot 1999$ ways to place the blue balls. But htere are $3!=6$ different placements, which produce the same partition of 2000 into three different summands, and $\frac{3!}{2}=3$ different placements, which produce the same partition of 2000 into summands two which are equal. Thus $6 x+3 y=$ 1999.999. But $y=999$, because the number appering twice in the partition can be any of the numbers $1,2, \ldots, 999$. This leads to $x=998 \cdot 333$, so $x+y=1001 \cdot 333=333333$.

333333

Combinatorics

math-word-problem

Yes

Yes

olympiads

01.3. Determine the number of real roots of the equation $$ x^{8}-x^{7}+2 x^{6}-2 x^{5}+3 x^{4}-3 x^{3}+4 x^{2}-4 x+\frac{5}{2}=0 $$

Solution. Write $$ \begin{gathered} x^{8}-x^{7}+2 x^{6}-2 x^{5}+3 x^{4}-3 x^{3}+4 x^{2}-4 x+\frac{5}{2} \\ =x(x-1)\left(x^{6}+2 x^{4}+3 x^{2}+4\right)+\frac{5}{2} \end{gathered} $$ If $x(x-1) \geq 0$, i.e. $x \leq 0$ or $x \geq 1$, the equation has no roots. If $0x(x-1)=\left(x-\frac{1}{2}\right)^{2}-\frac{1}{4} \geq-\frac{1}{4}$ and $x^{6}+2 x^{4}+3 x+4<1+2+3+4=10$. The value of the left-hand side of the equation now is larger than $-\frac{1}{4} \cdot 10+\frac{5}{2}=0$. The equation has no roots in the interval $(0,1)$ either.

0

Algebra

math-word-problem

Yes

Yes

olympiads

05.1. Find all positive integers $k$ such that the product of the digits of $k$, in the decimal system, equals $$ \frac{25}{8} k-211 $$

Solution. Let $$ a=\sum_{k=0}^{n} a_{k} 10^{k}, \quad 0 \leq a_{k} \leq 9, \text { for } 0 \leq k \leq n-1,1 \leq a_{n} \leq 9 $$ Set $$ f(a)=\prod_{k=0}^{n} a_{k} $$ Since $$ f(a)=\frac{25}{8} a-211 \geq 0 $$ $a \geq \frac{8}{25} \cdot 211=\frac{1688}{25}>66$. Also, $f(a)$ is an integer, and $\operatorname{gcf}(8,25)=1$, so $8 \mid a$. On the other hand, $$ f(a) \leq 9^{n-1} a_{n} \leq 10^{n} a_{n} \leq a $$ So $$ \frac{25}{8} a-211 \leq a $$ or $a \leq \frac{8}{17} \cdot 211=\frac{1688}{17}<100$. The only multiples of 8 between 66 and 100 are $72,80,88$, and 96. Now $25 \cdot 9-211=17=7 \cdot 2,25 \cdot 10-211=39 \neq 8 \cdot 0,25 \cdot 11-211=64=8 \cdot 8$, and $25 \cdot 12-211=89 \neq 9 \cdot 6$. So 72 and 88 are the numbers asked for.

7288

Number Theory

math-word-problem

Yes

Yes

olympiads

06.3. A sequence of positive integers $\left\{a_{n}\right\}$ is given by $$ a_{0}=m \quad \text { and } \quad a_{n+1}=a_{n}^{5}+487 $$ for all $n \geq 0$. Determine all values of $m$ for which the sequence contains as many square numbers as possible.

Solution. Consider the expression $x^{5}+487$ modulo 4. Clearly $x \equiv 0 \Rightarrow x^{5}+487 \equiv 3$, $x \equiv 1 \Rightarrow x^{5}+487 \equiv 0 ; x \equiv 2 \Rightarrow x^{5}+487 \equiv 3$, and $x \equiv 3 \Rightarrow x^{5}+487 \equiv 2$. Square numbers are always $\equiv 0$ or $\equiv 1 \bmod 4$. If there is an even square in the sequence, then all subsequent numbers of the sequence are either $\equiv 2$ or $\equiv 3 \bmod 4$, and hence not squares. If there is an odd square in the sequence, then the following number in the sequence can be an even square, but then none of the other numbers are squares. So the maximal number of squares in the sequence is two. In this case the first number of the sequence has to be the first square, since no number of the sequence following another one satisfies $x \equiv 1 \bmod 4$. We have to find numbers $k^{2}$ such that $k^{10}+487=n^{2}$. We factorize $n^{2}-k^{10}$. Because 487 is a prime, $n-k^{5}=1$ and $n+k^{5}=487$ or $n=244$ and $k=3$. The only solution of the problem thus is $m=3^{2}=9$.

9

Number Theory

math-word-problem

Yes

Yes

olympiads

09.2. On a faded piece of paper it is possible, with some effort, to discern the following: $$ \left(x^{2}+x+a\right)\left(x^{15}-\ldots\right)=x^{17}+x^{13}+x^{5}-90 x^{4}+x-90 $$ Some parts have got lost, partly the constant term of the first factor of the left side, partly the main part of the other factor. It would be possible to restore the polynomial forming the other factor, but we restrict ourselves to asking the question: What is the value of the constant term a? We assume that all polynomials in the statement above have only integer coefficients.

Solution. We denote the polynomial $x^{2}+x+a$ by $P_{a}(x)$, the polynomial forming the other factor of the left side by $Q(x)$ and the polynomial on the right side by $R(x)$. The polynomials are integer valued for every integer $x$. For $x=0$ we get $P_{a}(0)=a$ and $R(0)=-90$, so $a$ is a divisor of $90=2 \cdot 3 \cdot 3 \cdot 5$. For $x=-1$ we get $P_{a}(-1)=-184$, so $a$ is also a divisor of $184=2 \cdot 2 \cdot 2 \cdot 23$. But the only prime factor in common is 2 . So the only possibilities for $a$ are $\pm 2$ and $\pm 1$. If $a=1$, we get for $x=1$ that $P_{1}(1)=3$, while $R(1)=4-180=-176$, which cannot be divided by 3 . If $a=-2$ we get for $x=1$ that $P_{2}(1)=0$, i.e. the left side is equal to 0 , while the right side is equal to $R(1)=-176$, which is different from 0 . Neither $a=1$ nor $a=-2$ will thus work. It remains to check $a=2$ and $a=-1$. Before we use the procedure above again, we need a factorization of $R(x)$. We observe that $x^{4}+1$ is a divisor of $R(x)$, since the right side may be written as $\left(x^{4}+1\right)\left(x^{13}+x-90\right)$. If $a=-1$ we get for $x=2$ that $P_{1}(2)=5$, while $x^{4}+1=17$ and $x^{13}+x-90=8104$. So the right hand side is not divisible by 5 . Now, the only remaining possibility is $a=2$, i.e. $x^{2}+x+2$ is a divisor of $R(x)$.

2

Algebra

math-word-problem

Yes

Yes

olympiads

10.3. Laura has 2010 lamps connected with 2010 buttons in front of her. For each button, she wants to know the corresponding lamp. In order to do this, she observes which lamps are lit when Richard presses a selection of buttons. (Not pressing anything is also a possible selection.) Richard always presses the buttons simultaneously, so the lamps are lit simultaneously, too. a) If Richard chooses the buttons to be pressed, what is the maximum number of different combinations of buttons he can press until Laura can assign the buttons to the lamps correctly? b) Supposing that Laura will choose the combinations of buttons to be pressed, what is the minimum number of attempts she has to do until she is able to associate the buttons with the lamps in a correct way?

Solution. a) Let us say that two lamps are separated, if one of the lamps is turned on while the other lamp remains off. Laura can find out which lamps belong to the buttons if every two lamps are separated. Let Richard choose two arbitrary lamps. To begin with, he turns both lamps on and then varies all the other lamps in all possible ways. There are $2^{2008}$ different combinations for the remaining $2010-2=2008$ lamps. Then Richard turns the two chosen lamps off. Also, at this time there are $2^{2008}$ combinations for the remaining lamps. Consequently, for the $2^{2009}$ combinations in all, it is not possible to separate the two lamps of the first pair. However, we cannot avoid the separation if we add one more combination. Indeed, for every pair of lamps, we see that if we turn on a combination of lamps $2^{2009}+1$ times, there must be at least one setup where exactly one of the lamps is turned on and the other is turned off. Thus, the answer is $2^{2009}+1$. b) For every new step with a combination of lamps turned on, we get a partition of the set of lamps into smaller and smaller subsets where elements belonging to the same subset cannot be separated. In each step every subset is either unchanged or divided into two smaller parts, i.e. the total number of subsets after $\mathrm{k}$ steps will be at most $2^{k}$. We are finished when the number of subsets is equal to 2010 , so the answer is at least $\left\lceil\log _{2} 2010\right\rceil=11$. But it is easy to see that Laura certainly can choose buttons in every step in such a way that there are at most $2^{11-k}$ lamps in every part of the partition after $k$ steps. Thus, the answer is 11 .

11

Combinatorics

math-word-problem

Yes

Yes

olympiads

10.4. A positive integer is called simple if its ordinary decimal representation consists entirely of zeroes and ones. Find the least positive integer $k$ such that each positive integer $n$ can be written as $n=a_{1} \pm a_{2} \pm a_{3} \pm \cdots \pm a_{k}$, where $a_{1}, \ldots, a_{k}$ are simple.

Solution. We can always write $n=a_{l}+a_{2}+\cdots+a_{9}$ where $a_{j}$ has 1 's in the places where $n$ has digits greater or equal to $j$ and 0 's in the other places. So $k \leq 9$. To show that $k \geq 9$, consider $n=10203040506070809$. Suppose $n=a_{l}+a_{2}+\cdots+a_{j}-a_{j+l}-a_{j+2}-\cdots-a_{k}$, where $a_{l}, \ldots, a_{k}$ are simple, and $k<9$. Then all digits of $b_{l}=a_{l}+\cdots+a_{j}$ are not greater than $j$ and all digits of $b_{2}=a_{j+l}+\cdots+a_{k}$ are not greater than $k-j$. We have $n+b_{2}=b_{l}$. We perform column addition of $n$ and $b_{2}$ and consider digit $j+1$ in the number $n$. There will be no carry digit coming from lower decimal places, since the sum there is less that $10 \ldots 0+88 \ldots 8=98 \ldots 8$. So in the column of $j+1$ we get the sum of $j+1$ and the corresponding digit in $b_{2}$. The resulting digit should be less than $j+1$. Thus in the corresponding place in $b_{2}$ we have at least $9-j$. But $9-j \leq k-j$, implying $k \geq 9$. Hence, we have proved that the maximal $k$ is 9 .

9

Number Theory

math-word-problem

Yes

Yes

olympiads

## Problem 2 Let $A B C D$ be a cyclic quadrilateral satisfying $A B=A D$ and $A B+B C=C D$. Determine $\angle C D A$.

Solution 2 Answer: $\angle C D A=60^{\circ}$. Choose the point $E$ on the segment $C D$ such that $D E=A D$. Then $C E=C D-A D=$ $C D-A B=B C$, and hence the triangle $C E B$ is isosceles.  Now, since $A B=A D$ then $\angle B C A=\angle A C D$. This shows that $C A$ is the bisector of $\angle B C D=\angle B C E$. In an isosceles triangle, the bisector of the apex angle is also the perpendicular bisector of the base. Hence $A$ is on the perpendicular bisector of $B E$, and $A E=A B=A D=D E$. This shows that triangle $A E D$ is equilateral, and thus $\angle C D A=60^{\circ}$.

60

Geometry

math-word-problem

Yes

Yes

olympiads

## Problem 4 King George has decided to connect the 1680 islands in his kingdom by bridges. Unfortunately the rebel movement will destroy two bridges after all the bridges have been built, but not two bridges from the same island. What is the minimal number of bridges the King has to build in order to make sure that it is still possible to travel by bridges between any two of the 1680 islands after the rebel movement has destroyed two bridges?

Solution 4 Answer: 2016 An island cannot be connected with just one bridge, since this bridge could be destroyed. Consider the case of two islands, each with only two bridges, connected by a bridge. (It is not possible that they are connected with two bridges, since then they would be isolated from the other islands no matter what.) If they are also connected to two separate islands, then they would be isolated if the rebel movement destroys the two bridges from these islands not connecting the two. So the two bridges not connecting them must go to the same island. That third island must have at least two other bridges, otherwise the rebel movement could cut off these three islands. Suppose there is a pair of islands with exactly two bridges that are connected to each other. From the above it is easy to see that removing the pair (and the three bridges connected to them) must leave a set of islands with the same properties. Continue removing such pairs, until there are none left. (Note that the reduced set of islands could have a new such pair and that also needs to be removed.) Suppose we are left with $n$ islands and since two islands are removed at a time, $n$ must be an even number. And from the argument above it is clear that $n \geq 4$. Consider the remaining set of islands and let $x$ be the number of islands with exactly two bridges (which now are not connected to each other). Then $n-x$ islands have at least three bridges each. Let $B^{\prime}$ be the number of bridges in the reduced set. Now $B^{\prime} \geq 2 x$ and $2 B^{\prime} \geq 2 x+3(n-x)=3 n-x$. Hence $2 B^{\prime} \geq \max (4 x, 3 n-x) \geq 4 \cdot \frac{3 n}{5}$, and thus $B^{\prime} \geq \frac{6 n}{5}$. Now let $B$ be the number of bridges in the original set. Then $$ B=B^{\prime}+3 \cdot \frac{1680-n}{2} \geq \frac{6 n}{5}+\frac{6(1680-n)}{4} \geq \frac{6 \cdot 1680}{5}=2016 $$ It is possible to construct an example with exactly 2016 bridges: Take 672 of the islands and number them $0,1,2, \ldots 671$. Connect island number $i$ with the islands numbered $i-1$, $i+1$ and $i+336$ (modulo 672). This gives 1008 bridges. We now have a circular path of 672 bridges: $0-1-2-\cdots-671-0$. If one of these 672 bridges are destroyed, the 672 islands are still connected. If two of these bridges are destroyed, the path is broken into two parts. Let $i$ be an island on the shortest path (if they have the same length, just pick a random one). Then island $i+336$ (modulo 672) must be on the other part of the path, and the bridge connecting these two islands will connect the two paths. Hence no matter which two bridges the rebel movement destroys, it is possible to travel between any of the 672 islands. Now for every of the 1008 bridges above, replace it with two bridges with a new island between the two. This increases the number of bridges to 2016 and the number of islands to $672+1008=1680$ completing the construction. Since the rebel movement does not destroy two bridges from the same island, the same argument as above shows that with this construction it is possible to travel between any of the 1680 islands after the destruction of the two bridges.

2016

Combinatorics

math-word-problem

Yes

Yes

olympiads

PROBLEM 1. The real numbers $a, b, c$ are such that $a^{2}+b^{2}=2 c^{2}$, and also such that $a \neq b, c \neq-a, c \neq-b$. Show that $$ \frac{(a+b+2 c)\left(2 a^{2}-b^{2}-c^{2}\right)}{(a-b)(a+c)(b+c)} $$ is an integer.

SolUTiON. Let us first note that $$ \frac{a+b+2 c}{(a+c)(b+c)}=\frac{(a+c)+(b+c)}{(a+c)(b+c)}=\frac{1}{a+c}+\frac{1}{b+c} $$ Further we have $$ 2 a^{2}-b^{2}-c^{2}=2 a^{2}-\left(2 c^{2}-a^{2}\right)-c^{2}=3 a^{2}-3 c^{2}=3(a+c)(a-c) $$ and $$ 2 a^{2}-b^{2}-c^{2}=2\left(2 c^{2}-b^{2}\right)-b^{2}-c^{2}=3 c^{2}-3 b^{2}=3(b+c)(c-b) $$ so that $$ \frac{(a+b+2 c)\left(2 a^{2}-b^{2}-c^{2}\right)}{(a-b)(a+c)(b+c)}=\frac{3(a-c)+3(c-b)}{a-b}=\frac{3(a-b)}{a-b}=3 $$ an integer.

3

Algebra

proof

Yes

Yes

olympiads

Problem 3. Find the smallest positive integer $n$, such that there exist $n$ integers $x_{1}, x_{2}, \ldots, x_{n}$ (not necessarily different), with $1 \leq x_{k} \leq n, 1 \leq k \leq n$, and such that $$ x_{1}+x_{2}+\cdots+x_{n}=\frac{n(n+1)}{2}, \quad \text { and } \quad x_{1} x_{2} \cdots x_{n}=n! $$ but $\left\{x_{1}, x_{2}, \ldots, x_{n}\right\} \neq\{1,2, \ldots, n\}$.

Solution. If it is possible to find a set of numbers as required for some $n=k$, then it will also be possible for $n=k+1$ (choose $x_{1}, \ldots, x_{k}$ as for $n=k$, and let $x_{k+1}=k+1$ ). Thus we have to find a positive integer $n$ such that a set as required exists, and prove that such a set does not exist for $n-1$. For $n=9$ we have $8+6+3=9+4+4$, and $8 \cdot 6 \cdot 3=9 \cdot 4 \cdot 4$, so that a set of numbers as required will exist for all $n \geq 9$. It remains to eliminate $n=8$. Assume $x_{1}, \ldots, x_{8}$ are numbers that satisfy the conditions of the problem. Since 5 and 7 are primes, and since $2 \cdot 5>8$ and $2 \cdot 7>8$, two of the $x$-numbers have to be equal to 5 and 7 ; without loss of generality we can assume that $x_{1}=5, x_{2}=7$. For the remaining numbers we have $x_{3} x_{4} \cdots x_{8}=2^{7} \cdot 3^{2}$, and $x_{3}+x_{4}+\cdots+x_{8}=36-12=24$. Since $3^{2}=9>8$, it follows that exactly two of the numbers $x_{3}, \ldots, x_{8}$ are divisible by 3 , and the rest of the numbers are powers of 2. There are three possible cases to consider: two of the numbers are equal to 3 ; two of the numbers are equal to 6 ; one number is equal to 3 and another one is equal to 6 . Case 1. $x_{3}=x_{4}=3$ We then have $x_{5}+x_{6}+x_{7}+x_{8}=18$, and $x_{5} x_{6} x_{7} x_{8}=2^{7}$. The possible powers of 2 with sum 18 are $(1,1,8,8)$ and $(2,4,4,8)$, none of them gives the product $2^{7}$. Case 2. $x_{3}=3, x_{4}=6$ We have $x_{5}+x_{6}+x_{7}+x_{8}=15$, and $x_{5} x_{6} x_{7} x_{8}=2^{6}$. It is immediate to check that the only possibility for the remaining numbers is $(1,2,4,8)$, which is not allowed, since it gives $\left\{x_{1}, x_{2}, \ldots, x_{8}\right\}=\{1,2, \ldots, 8\}$. Case 3. $x_{3}=x_{4}=6$ Now we have $x_{5}+x_{6}+x_{7}+x_{8}=12$, and $x_{5} x_{6} x_{7} x_{8}=2^{5}$. The possible powers of 2 which give the correct sum are $(1,1,2,8)$ and $(2,2,4,4)$, but again, they do not give the desired product. Thus the smallest positive integer with the required property is 9 .

9

Number Theory

math-word-problem

Yes

Yes

olympiads

XXI OM - II - Task 5 Given is the polynomial $ P(x) = \frac{1}{2} - \frac{1}{3}x + \frac{1}{6}x^2 $. Let $ Q(x) = \sum_{k=0}^{m} b_k x^k $ be a polynomial defined by the formula Calculate $ \sum_{k=0}^m |b_k| $.

We will use the following obvious facts: (1) The product of polynomials with positive coefficients is a polynomial with positive coefficients. (2) The sum of the coefficients of a polynomial $ f(x) $ is equal to $ f(1) $. (3) The absolute values of the corresponding coefficients of the polynomials $ f(x) $ and $ f(-x) $ are equal. Since the polynomial $ P(-x) = \frac{1}{2} + \frac{1}{3} x + \frac{1}{6} x^2 $ has positive coefficients, then based on (1) the polynomial $ R(x) = Q(-x) = P(-x) \cdot P(- x^3) \cdot P(- x^9) \cdot P(- x^{27}) \cdot P(- x^{81}) $ has positive coefficients. Based on (2) the sum of the coefficients of the polynomial $ R (x) $ is equal to $ R(1) = P(- 1) \cdot P(- 1) \cdot P(- 1) \cdot P(- 1) \cdot P(-1) = 1 $. Therefore, based on (3) the sum of the absolute values of the coefficients of the polynomial $ Q(x) $ is equal to the sum of the coefficients of the polynomial $ Q(- x) = R(x) $, which is $ 1 $.

1

Algebra

math-word-problem

Incomplete

Yes

olympiads

XXIV OM - I - Problem 10 Find the smallest natural number $ n > 1 $ with the following property: there exists a set $ Z $ consisting of $ n $ points in the plane such that every line $ AB $ ($ A, B \in Z $) is parallel to some other line $ CD $ ($ C, D \in Z $).

We will first prove that the set $ Z $ of vertices of a regular pentagon has the property given in the problem, that is, $ n \leq 5 $. We will show that each side of the regular pentagon is parallel to a certain diagonal and vice versa, each diagonal is parallel to a corresponding side. It suffices to prove that $ AB \parallel CE $ (Fig. 11). Since a circle can be circumscribed around quadrilateral $ ABCE $ (namely, it is the circle circumscribed around the given regular pentagon), we have $ \measuredangle A + \measuredangle BCE = \pi $. Since $ \measuredangle A = \measuredangle B $, it follows that $ \measuredangle BCE = \pi - \measuredangle B $, which proves that $ AB \parallel CE $. On the other hand, from the conditions of the problem, it follows that $ n \geq 4 $, since there are at least two different parallel lines, each containing at least two points of the set $ Z $. If $ n = 4 $ and points $ A, B, C, D $ satisfied the conditions of the problem, they would be the vertices of a trapezoid. However, none of the diagonals of a trapezoid is parallel to another line determined by its vertices. Therefore, $ n > 4 $, and from the previously proven inequality $ n \leq 5 $, it follows that $ n = 5 $.

5

Combinatorics

math-word-problem

Yes

Yes

olympiads

XXV - I - Task 1 During World War I, a battle took place near a certain castle. One of the shells destroyed a statue of a knight with a spear standing at the entrance to the castle. This happened on the last day of the month. The product of the day of the month, the month number, the length of the spear expressed in feet, half the age of the battery commander firing at the castle expressed in years, and half the time the statue stood expressed in years equals 451,066. In which year was the statue erected?

The last day of the month can only be $28$, $29$, $30$, or $31$. Of these numbers, only $29$ is a divisor of the number $451,066 = 2 \cdot 7 \cdot 11 \cdot 29 \cdot 101$. Therefore, the battle took place on February $29$ in a leap year. During World War I, only the year $1916$ was a leap year. From the problem statement, it follows that a divisor of the number $7 \cdot 11 \cdot 101$ is half the age of the battery commander. Only the number $11$ satisfies this condition, so the battery commander was $22$ years old. The length of the pike is a divisor of the number $7 \cdot 101$. Therefore, the pike was $7$ feet long. Thus, half the time the statue stood is $101$ years. It was erected $202$ years before the year $1916$, i.e., in the year $1714$.

1714

Logic and Puzzles

math-word-problem

Yes

Yes

olympiads

For a natural number $ k \geq 1 $, let $ p(k) $ denote the smallest prime number that is not a divisor of $ k $. If $ p(k) > 2 $, then we define $ q(k) $ as the product of all prime numbers less than $ p(k) $; if $ p(k) = 2 $, then we set $ q(k) = 1 $. We define the sequence $ (x_n) $ by the formulas: Determine all natural numbers $ n $ for which the equality $ x_n = 111111 $ holds.

Let $ p_0 = 2 $, $ p_1 = 3 $, $ p_2 = 5 $, $ p_3 = 7 $, $ \ldots $ be the increasing sequence of all prime numbers. The terms of the sequence $(x_n)$ under consideration can be characterized by the digits of the binary representation of the index $ n $. Specifically, for $ n \geq 1 $, the implication holds: For $ n = 1 $, this is true ($ x_1 = x_0 p(x_0)/q(x_0) = p(1)/q(1) = 2 = p_0 $). Assume the statement (*) is true for some natural number $ n \geq 1 $. If $ n $ is even, then $ c_0 = 0 $, so Thus, $ p(x_n) = 2 $, $ q(x_n) = 1 $, and according to the given recursive formula: $ x_{n+1} = 2x_n $. Therefore, the rule (*) remains valid for $ n + 1 $. If, on the other hand, $ n $ is odd, then $ c_0 = 1 $. Assume the binary representation of $ n $ does not consist entirely of ones (we will consider the case where it does shortly). Let $ j $ be the smallest index for which $ c_j = 0 $. This means The rule (*) (which holds for the number $ n $ by the induction hypothesis) gives the equality Thus, $ p(x_n) = p_j $, $ q(x_n) = \prod_{i=0}^{j-1} p_i $. The numbers $ n+1 $ and $ x_{n+1} $ then have the representations: Therefore, in this case as well, the rule (*) is valid for $ n + 1 $. The remaining case to consider is when the binary representation of $ n $ consists entirely of ones: According to the induction hypothesis (*) and the given recursive formula, we have: Thus, As we can see, in this case as well, the rule (*) remains valid for $ n + 1 $. By the principle of induction, the implication (*) holds for all natural numbers $ n \geq 1 $. Different numbers $ n $ have different binary expansions. It follows from the characterization (*) that no number repeats in the sequence $(x_n)$. It remains to note that in the prime factorization of the number $ 111111 $, we have $ 111111 = 3 \cdot 7 \cdot 11 \cdot 13 \cdot 37 $, which satisfies the equation $ 111111 = p_1 p_3 p_4 p_5 p_{11} $. Therefore, for the number $ n = 2^1 + 2^3 + 2^4 + 2^5 + 2^{11} = 2106 $, and only for it, the equality $ x_n = 111111 $ holds. Note: From the reasoning above, it follows that the sequence $(x_n)$ contains all positive square-free integers, i.e., integers that are not divisible by the square of any prime number; in other words, integers that have a prime factorization into distinct prime factors; and each such number appears in the sequence exactly once. The recursive formula generating this sequence mimics the algorithm for adding one in the binary positional system. Noticing this fact and its precise description is the essence of the solution above.

2106

Number Theory

math-word-problem

Incomplete

Yes

olympiads

LVII OM - I - Problem 4 Participants in a mathematics competition solved six problems, each graded with one of the scores 6, 5, 2, 0. It turned out that for every pair of participants $ A, B $, there are two problems such that in each of them $ A $ received a different score than $ B $. Determine the maximum number of participants for which such a situation is possible.

We will show that the largest number of participants for which such a situation is possible is 1024. We will continue to assume that the permissible ratings are the numbers 0, 1, 2, 3 (instead of 5 points, we give 4, and then divide each rating by 2). Let $ P = \{0,1,2,3\} $ and consider the set Set $ X $ obviously has 4096 elements. We will consider subsets $ A $ of set $ X $ with the following property (*): (*) If $ (a_1,a_2,\dots,a_6) $, $ (b_1,b_2,\dots,b_6) \in A $, then there exist $i, j$ such that It suffices to show that the largest number of elements in set $ A $ with property (*) is 1024. First, we show that if set $ A $ has property (*), then it has at most 1024 elements. Assume, therefore, that we have a subset $ A $ of set $ X $ with property (*) and suppose that it has at least 1025 elements. Since there are exactly 1024 sequences of length 5 with terms from the four-element set $ P $, it follows from the pigeonhole principle that in set $ A $ there are at least two sequences that have the same terms from the first to the fifth. These sequences differ, therefore, only in one term—the sixth, which contradicts property (*). Therefore, set $ A $ has at most 1024 elements. Now we show that there exists a set $ A $ with at least 1024 elements and having property (*). It suffices to take the following set: First, we show that set $ A $ has at least 1024 elements. Take any numbers $ a_1,a_2,\dots,a_5 \in P $. We can make such a choice in 1024 ways. Let $ r $ be the remainder of the division of the sum $ a_1+a_2+\dots+a_5 $ by 4, and let $ a_6 = 4 - r $. Then, of course, $ (a_1,a_2,\dots,a_6) \in A $, so we have indicated at least 1024 different sequences in set $ A $. Finally, we show that set $ A $ has property (*). Suppose that and sequences $ (a_1,a_2,\dots,a_6) $ and $ (b_1,b_2,\dots,b_6) $ differ in only one term, say the term with index $ k: \; a_k \neq b_k $, where $ 1 \leq k \leq 6 $ and $ a_i = b_i $ for $ i \neq k $. Since the numbers $ a_1 +a_2 +\dots +a_6 $ and $ b_1 +b_2 +\dots+b_6 $ are divisible by 4, their difference is also divisible by 4. But Thus, the number $ a_k - b_k $ is divisible by 4. Since $ a_k, b_k \in P $, then In the set $ \{-3,-2, -1,0,1,2,3\} $, there is only one number divisible by 4, namely 0. Therefore, $ a_k = b_k $, contrary to the assumption that sequences $ (a_1,a_2,\dots,a_6) $ and $ (b_1,b_2,\dots,b_6) $ differ in the term with index $ k $. This contradiction proves that set $ A $ has property (*), which completes the proof.

1024

Combinatorics

math-word-problem

Yes

Yes

olympiads

XXXI - III - Task 3 We choose an integer $ k $ from the interval $ [1, 99] $. Then we toss a coin 100 times (the outcomes of the tosses are independent). Let Denote by $ M_k $ the event that there exists a number $ i $ such that $ k + \varepsilon_1 + \ldots + \varepsilon_i = 100 $. What should $ k $ be so that the probability of the event $ M_k $ is maximized?

Let $ P(M_k) $ be the probability of the event $ M_k $. Of course, $ P(M_{99})=\frac{1}{2} $, $ p(M_{98})=\frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4} $. We will show that for $ k < 98 $, $ P(M_k) < \frac{3}{4} $. First, notice that Indeed, if a head (probability of this is $ \frac{1}{2} $) comes up on the first toss, the favorable event for $ M_{k-2} $ is that $ (k - 2) + 1 + \varepsilon_2 + \ldots + \varepsilon_i = 100 $, so $ k - 1 + \varepsilon_2 +\ldots + \varepsilon_i = 100 $. If, however, a tail comes up on the first toss, the favorable event is that $ k + \varepsilon_2 + \ldots + \varepsilon_i = 100 $. Using (1), we can prove by induction that For $ n = 1 $, by (1) we have Assume that for some $ n $ ($ 1 < n < 97 $) and for $ k = 1, 2, \ldots, n $, $ P(M_{98-k}) < \frac{3}{4} $. It follows from the principle of induction that $ P(M_k) < \frac{3}{4} $ for $ 1 \leq k < 98 $. Since moreover $ P(M_{99}) < \frac{3}{4} $, the probability of the event $ M_k $ is the greatest for $ k = 98 $.

98

Combinatorics

math-word-problem

Incomplete

Yes

olympiads

XXIII OM - I - Problem 7 A broken line contained in a square with a side length of 50 has the property that the distance from any point of this square to it is less than 1. Prove that the length of this broken line is greater than 1248.

Let the broken line $ A_1A_2 \ldots A_n $ have the property given in the problem. Denote by $ K_i $ ($ i= 1, 2, \ldots, n $) the circle with center at point $ A_i $ and radius of length $ 1 $, and by $ F_i $ ($ i= 1, 2, \ldots, n-1 $) the figure bounded by segments parallel to segment $ \overline{A_iA_{i+1}} $ and at a distance of $ 1 $ from it, as well as by arcs of circles $ K_i $ and $ K_{i+1} $ (in Fig. 5, the figure $ F_i $ is shaded). The set of points in the plane at a distance of less than $ 1 $ from some point of segment $ \overline{A_i A_{i+1}} $ is contained in the union of circles $ K_i $ and $ K_{i+1} $ and figure $ F_i $. From the conditions of the problem, it follows that the given square is contained in the set The area of the figure $ K_i \cup F_i $ is not less than $ 2A_iA_{i+1} $ (Fig. 5), and the area of the circle $ K_n $ is equal to $ \pi $. Therefore, the area of the given square does not exceed the sum of the areas of these figures, i.e., Hence the length of the broken line $ \displaystyle = \sum_{i=1}^{n-1} A_iA_{i+1} \geq 1250 - \frac{\pi}{2} > 1248 $.

1248

Geometry

proof

Yes

Yes

olympiads

LVII OM - III - Problem 2 Determine all positive integers $ k $ for which the number $ 3^k+5^k $ is a power of an integer with an exponent greater than 1.

If $ k $ is an even number, then the numbers $ 3^k $ and $ 5^k $ are squares of odd numbers, giving a remainder of 1 when divided by 4. Hence, the number $ 3^k + 5^k $ gives a remainder of 2 when divided by 4, and thus is divisible by 2 but not by $ 2^2 $. Such a number cannot be a power of an integer with an exponent greater than 1. If $ k $ is an odd number, then The second factor on the right side of the above relationship contains an odd (equal to $ k $) number of odd summands. Hence, the number $ 3^k + 5^k $ is divisible by 8 and not by 16. If this number is a power of an integer with an exponent greater than 1, then it must be a cube of an integer. If $ k = 1 $, then the considered number is a cube of an integer: $ 3^1 + 5^1 = 2^3 $. Let us assume in the further part of the reasoning that $ k \geq 3 $. From the relationship it follows that cubes of integers give remainders of 0, 1, 8 when divided by 9. For $ k \geq 3 $ we have $ 9 | 3^k $, so $ 3^k + 5^k \equiv 5^k (\mod 9) $. The remainders of the numbers $ 5, 5^2, 5^3, 5^4, 5^5, 5^6 $ when divided by 9 are 5, 7, 8, 4, 2, 1, respectively. Therefore, if $ 3^k + 5^k $ is a cube of an integer for $ k \geq 3 $, then $ 3 | k $. We have previously shown that $ k $ cannot be an even number. Thus, the number $ k $ is of the form $ 6l + 3 $, where $ l $ is a non-negative integer. From the relationship $ 3^3 \equiv 5^3 \equiv 6 (\mod 7) $ and $ 3^6 \equiv 5^6 \equiv 1 (\mod 7) $, it follows that However, from direct verification, we obtain that the cube of an integer gives a remainder of 0, 1, or 6 when divided by 7: Therefore, the considered number for $ k \geq 3 $ cannot be a cube of an integer, which concludes the solution of the problem. Answer: $ k = 1 $.

1

Number Theory

math-word-problem

Yes

Yes

olympiads

XLVIII OM - III - Problem 1 Positive integers $ x_1 $, $ x_2 $, $ x_3 $, $ x_4 $, $ x_5 $, $ x_6 $, $ x_7 $ satisfy the conditions: Determine $ x_7 $.

Let's denote: $ a = x_1+x_2 $, $ b = x_3 $, $ c = x_2+x_3 $. Then $ x_2 = c - b $, $ x_1 = a-x_2=a+b-c $. These are positive numbers by assumption. Thus Further, $ x_4 = ab, x_5 = abc, x_6 = abc(ab + b) = a(a + 1)b^2c $. From inequality (1), it follows that $ a > 1 $. The only divisors of the number $ 144 $ that have the form $ a(a + 1) $ (where $ a > 1 $) are the numbers $ 6 $, $ 12 $, $ 72 $, corresponding to the values $ a = 2 $, $ a = 3 $, $ a = 8 $. The product $ b^2c $ should have the respective values: $ 24 $, $ 12 $, $ 2 $. The first possibility leads to a contradiction with condition (1). The second and third possibilities, combined with condition (1), yield the solutions $ (a,b,c) = (3,2,3) $ and $ (a,b,c) = (8,1,2) $. In each case, $ x_7 = x_6(x_5 + x_4) = 144ab(c+1) = 144 \cdot 24 = 3456 $.

3456

Number Theory

math-word-problem

Incomplete

Yes

olympiads

VIII OM - I - Task 6 Find a four-digit number, whose first two digits are the same, the last two digits are the same, and which is a square of an integer.

If $ x $ is the number sought, then where $ a $ and $ b $ are integers satisfying the inequalities $ 0 < a \leq 9 $, $ 0 \leq b \leq 9 $. The number $ x $ is divisible by $ 11 $, since Since $ x $ is a perfect square, being divisible by $ 11 $ it must be divisible by $ 11^2 $, so the number is divisible by $ 11 $. It follows that $ a + b $ is divisible by $ 11 $, and since $ 0 < a + b \leq 18 $, then $ a + b = 11 $. Therefore, from which we infer that $ 9a + 1 $ is the square of some natural number $ m $: Since $ 9a + 1 \leq 82 $, then $ m \leq 9 $. From the above, It follows from this equality that the product $ (m + 1) (m - 1) $ is divisible by $ 9 $, and since at most one of the numbers $ m + 1 $ and $ m - 1 $ is divisible by $ 3 $, then one of them is divisible by $ 9 $. Considering that the natural number $ m $ is less than $ 10 $, we conclude from this that $ m + 1 = 9 $, so $ m = 8 $. In this case, $ a = 7 $, $ b = 4 $, and the sought number is $ 7744 = (88)^2 $.

7744

Number Theory

math-word-problem

Yes

Yes

olympiads

LII OM - I - Problem 12 We consider sequences of integers $ x_0,x_1,\ldots,x_{2000} $ satisfying the conditions Find the smallest value of the expression $ |x_1 + x_2 +\ldots + x_{2000}| $.

Answer: The smallest value of the given expression is $ 12 $. We will prove by induction that for any sequence $ (x_n) $ satisfying the conditions of the problem, the following equality holds For $ n = 0 $, both sides of the above equality are equal to $ 0 $. So, assume that the relation (1) is true for some $ n $. Then which completes the inductive proof of formula (1). Therefore, The square number closest to $ 2001 $ is $ 2025 = 45^2 $. Hence, from the above equality, we conclude that the expression $ |x_1 + x_2 +\ldots + x_{2000}| $ cannot take a value less than $ \frac{1}{2}|2025 - 2001| =12 $. It remains to note that the value $ 12 $ can be achieved by taking The above sequence satisfies the conditions of the problem. Moreover, $ x_{2000} = 44 $, which, by equality (1), gives $ |x_1 + x_2 +\ldots + x_{2000}| = 12 $.

12

Number Theory

math-word-problem

Incomplete

Yes

olympiads

LI OM - II - Problem 4 Point $ I $ is the center of the circle inscribed in triangle $ ABC $, where $ AB \neq AC $. Lines $ BI $ and $ CI $ intersect sides $ AC $ and $ AB $ at points $ D $ and $ E $, respectively. Determine all possible measures of angle $ BAC $ for which the equality $ DI = EI $ can hold.

We will show that the only value taken by angle $ BAC $ is $ 60^\circ $. By the Law of Sines applied to triangles $ ADI $ and $ AEI $, we obtain $ \sin \measuredangle AEI = \sin \measuredangle ADI $. Hence, om51_2r_img_6.jpg First, suppose that the equality $ \measuredangle AEI = \measuredangle ADI $ holds (Fig. 1). Then also $ \measuredangle AIE = \measuredangle AID $, which means that triangles $ AEI $ and $ ADI $ are congruent (angle-side-angle criterion). Therefore, $ AD = AE $. This proves that triangles $ ADB $ and $ AEC $ are also congruent (angle-side-angle criterion). Hence, we obtain $ AB = AC $, which contradicts the assumptions made in the problem statement. om51_2r_img_7.jpg The remaining case to consider is when $ \measuredangle AEI + \measuredangle ADI = 180^\circ $ (Fig. 2). Then points $ A $, $ E $, $ I $, $ D $ lie on a single circle. Therefore, From this, we obtain which means $ \measuredangle BAC = 60^\circ $. To complete the solution, it remains to show that there exists a triangle $ ABC $ in which $ AB \neq AC $, $ \measuredangle BAC = 60^\circ $, and $ DI = EI $. We will show more: in any triangle $ ABC $ with $ \measuredangle BAC = 60^\circ $, the equality $ DI = EI $ holds. If $ \measuredangle BAC = 60^\circ $, then Therefore, a circle can be circumscribed around quadrilateral $ AEID $. Since $ AI $ is the angle bisector of $ \angle EAD $, then $ DI = EI $.

60

Geometry

math-word-problem

Yes

Yes

olympiads

XXXIV OM - II - Problem 6 For a given number $ n $, let $ p_n $ denote the probability that when a pair of integers $ k, m $ satisfying the conditions $ 0 \leq k \leq m \leq 2^n $ is chosen at random (each pair is equally likely), the number $ \binom{m}{k} $ is even. Calculate $ \lim_{n\to \infty} p_n $.

om34_2r_img_10.jpg The diagram in Figure 10 shows Pascal's triangle written modulo $2$, meaning it has zeros and ones in the places where the usual Pascal's triangle has even and odd numbers, respectively. Just like in the usual Pascal's triangle, each element here is the sum of the elements directly above it, according to the addition table modulo 2: $0+0=0$, $0+1=1$, $1+0=1$, $1+1 = 0$. The drawn horizontal lines divide this triangle into layers: - layer $W_0$ consisting of row number $0$, - layer $W_1$ consisting of row number $1$, - layer $W_2$ consisting of rows number $2$ and $3$, - layer $W_3$ consisting of rows number $4$, $5$, $6$, $7$, - $\ldots$ - layer $W_j$ consisting of rows with numbers from $2^{j-1}$ to $2^j-1$ - $\ldots$ We will prove by induction that in any layer $W_j$ there are two disjoint triangles, each identical to the triangle formed by the layers $W_0, W_1, \ldots, W_{j-1}$, and that these two triangles are separated by an inverted triangle composed entirely of zeros; the top row of the layer consists of extreme ones and zeros, and the bottom row of the layer consists of all ones. For $j = 1, 2, 3$, this is visible from the diagrams (the diagonal lines define the division into the triangles mentioned). Assume the above relationships are true for some $j$ and consider the layer $W_{j+1}$. By the induction hypothesis, the last row of layer $W_j$ has the form $111\ldots 111$, and thus, according to the addition rule given above, the first row of layer $W_{j+1}$ has the form $100\ldots 001$. The extreme ones of this row give rise to two new copies of Pascal's triangle, and between them are zeros. Layer $W_{j+1}$ has as many rows as the layers $W_0, W_1, \ldots, W_j$ taken together, so at the base of each of these two triangles at the bottom of layer $W_{j+1}$ will be the same pattern of digits as in the last row of layer $W_j$, i.e., a row of ones. They will fill the last row of layer $W_{j+1}$. This completes the proof of the inductive thesis. The number $p_n$ defined in the problem is the ratio of the number of zeros in the rows numbered from $0$ to $2^n$ to the number of all elements in these rows. These rows cover the layers $W_0, W_1, \ldots, W_n$ and the first row of layer $W_{n+1}$. From the relationship proved above, it follows that in the sum of layers $W_0, \ldots, W_j$ there are three times as many ones as in the sum of layers $W_0, \ldots, W_{j-1}\ (j = 1, 2, 3, \ldots)$, and since there is one one in layer $W_0$, there are $3^n$ ones in the sum of layers $W_0, \ldots, W_n$. Adding to this the two extreme ones of the next row, we see that there are $a_n = 3^n + 2$ ones in the considered fragment of Pascal's triangle. The total number of elements in this fragment is $b_n = 1 + 2 + 3 + \ldots + (2^n + 1) = (2^n + 1)(2^n + 2) / 2$, and thus there are $b_n - a_n$ zeros. The probability in question is \[ \frac{b_n - a_n}{b_n} = \frac{(2^n + 1)(2^n + 2) / 2 - (3^n + 2)}{(2^n + 1)(2^n + 2) / 2} \] Hence, \[ \lim_{n \to \infty} p_n = 1 \]

1

Combinatorics

math-word-problem

Yes

Yes

olympiads