Problems & Puzzles: Puzzles

Puzzle 116. A=B+C, A*B*C is a primorial...

Naohiro Nomoto asks for numbers A, B & C such that the following four rules apply:

1. A = B + C
2. A*B*C is a primorial
3. A, B, & C are, each, a product of n of distinct primes
4. gcd(A,B)=gcd(A,C)=1

Nomoto has found only two examples:

5 = 2 + 3

1105 = 231 + 874 [ 5*13*17 = 3*7*11 + 2*19*23 ]

Question 1: Find the next example

If the condition 3 is relaxed and we let that A, B & C are the product of distinct primes, not necessarily the same quantity, example:

10 = 3 + 7  [ 2*5 = 3 + 7 ]

There are some more examples, being the largest currently found, the following one:

496961=495726+1235 [ 17x23x31x41 = 2x3x7x11x29x37 + 5x13x19 ]

Question 2. Find the next 3 examples

Question 3. Can you argue if this sequence is finite or infinite?

___________
Notes:  All the numbers for the sequence described in the question 2 are a subset of the
Sloane's integers sequence A057035.

Hakan Summakoğlu wrote (May 2011)

Q2: 219965 = 213486 + 6479  [ 5x29x37x41 = 2x3x7x13x17x23 + 11x19x31]

***

Jan van Delden wrote (May 2011)

I was wondering about the number of solutions of Puzzle 116, I thought that could be improved. Must have missed this one..

My contribution:

Q1/Q2:

We have P:=p[n]#=a.b.c and a=b+c. Given P there are two equations with 3 unknowns. We can therefore (for instance) choose b and a and c follow. From this one can deduce that if b^2+4P/b is a square, say D^2, we have: a=(D+b)/2 and c=(D-b)/2.

Furthermore the minimum number of factors of P to be assigned to a can be calculated beforehand (using the largest prime factors of P), leading to the maximum number of factors to be assigned to b. Unfortunately this maximum is about n/3, which still gives a lot of tests to perform!

I found the following 20 solutions, in order of appearance (primarily Q2):

5=2+3 (Q1)

10=3+7 2x5=3+7

35=  2+33 5x7=2+3x11
22=  7+15 2x11=7+3x5
21=11+10 3x7=11+2x5

55=13+42 5x11=13+2x3x7

221=11+210 13x17=11+2x3x5x7
182=17+165 2x7x13=17+3x5x11
231=10+221 3x7x11=2x5+13x17

442=57+385  2x13x17=3x19+5x7x11
357=247+110 3x7x17=13x19+2x5x11

1430=119+1311 2x5x11x13=7x17+3x19x23
1105=874+231  5x13x17=2x19x23+3x7x11 (Q1)

2958=1495+1463  2x3x17x29=5x13x23+7x11x19

23345=374+22971 5x7x23x29=2x11x17+3x13x19x31

496961=1235+495726  17x23x31x41=5x13x19+2x3x7x11x29x37
219965=6479+213486  5x29x37x41=11x19x31+2x3x7x13x17x23
144210=16523+127687 2x3x5x11x19x23=13x31x41+7x17x29x37
226083=6118+219965  3x11x13x17x31=2x7x19x23+5x29x37x41
106981=57646+49335  7x17x29x31=2x19x37x41+3x5x11x13x23

No further solutions with n<=35

Q3:

I computed the minimum absolute distance between b^2+4P/b and the nearest square (where no solution was found), let's call this dist(D^2). The smallest distance 103 was found (n in [12,14-26]) at n=18, b=12509182. At first sight this is large, however a and c fail to be integer by a mere 0.000000133 (=dist(a),dist(b)). However the resulting distance between P and a.b.c is huge, since a.b.c=(D+b)/2.b.(D-b)/2=b.(D^2-b^2)/4, this gap is of the order b.dist(D^2)/4=12509182.103/4.

In general dist(a)=dist(b)=dist(D^2)/(4D) and dist(P)=b.dist(D^2)/4. And dist(D^2) is maximal D+1/4. More should be known about the distribution of dist(D^2) to say more.

***

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The text passage contains a series of mathematical puzzles and questions, along with some provided solutions and additional commentary from contributors.
multiple_questions_with_answers

1. the passage contains multiple mathematical questions (questions 1, 2, and 3).
2. answers to these questions are provided in the form of examples and solutions contributed by hakan summakoğlu and jan van delden.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
Problems & Puzzles: Puzzles

Puzzle 116. A=B+C, A*B*C is a primorial...

Naohiro Nomoto asks for numbers A, B & C such that the following four rules apply:

1. A = B + C
2. A*B*C is a primorial
3. A, B, & C are, each, a product of n of distinct primes
4. gcd(A,B)=gcd(A,C)=1

Nomoto has found only two examples:

5 = 2 + 3

1105 = 231 + 874 [ 5*13*17 = 3*7*11 + 2*19*23 ]

Question 1: Find the next example

If the condition 3 is relaxed and we let that A, B & C are the product of distinct primes, not necessarily the same quantity, example:

10 = 3 + 7  [ 2*5 = 3 + 7 ]

There are some more examples, being the largest currently found, the following one:

496961=495726+1235 [ 17x23x31x41 = 2x3x7x11x29x37 + 5x13x19 ]

Question 2. Find the next 3 examples

Question 3. Can you argue if this sequence is finite or infinite?

___________
Notes:  All the numbers for the sequence described in the question 2 are a subset of the
Sloane's integers sequence A057035.

Hakan Summakoğlu wrote (May 2011)

Q2: 219965 = 213486 + 6479  [ 5x29x37x41 = 2x3x7x13x17x23 + 11x19x31]

***

Jan van Delden wrote (May 2011)

I was wondering about the number of solutions of Puzzle 116, I thought that could be improved. Must have missed this one..

My contribution:

Q1/Q2:

We have P:=p[n]#=a.b.c and a=b+c. Given P there are two equations with 3 unknowns. We can therefore (for instance) choose b and a and c follow. From this one can deduce that if b^2+4P/b is a square, say D^2, we have: a=(D+b)/2 and c=(D-b)/2.

Furthermore the minimum number of factors of P to be assigned to a can be calculated beforehand (using the largest prime factors of P), leading to the maximum number of factors to be assigned to b. Unfortunately this maximum is about n/3, which still gives a lot of tests to perform!

I found the following 20 solutions, in order of appearance (primarily Q2):

5=2+3 (Q1)

10=3+7 2x5=3+7

35=  2+33 5x7=2+3x11
22=  7+15 2x11=7+3x5
21=11+10 3x7=11+2x5

55=13+42 5x11=13+2x3x7

221=11+210 13x17=11+2x3x5x7
182=17+165 2x7x13=17+3x5x11
231=10+221 3x7x11=2x5+13x17

442=57+385  2x13x17=3x19+5x7x11
357=247+110 3x7x17=13x19+2x5x11

1430=119+1311 2x5x11x13=7x17+3x19x23
1105=874+231  5x13x17=2x19x23+3x7x11 (Q1)

2958=1495+1463  2x3x17x29=5x13x23+7x11x19

23345=374+22971 5x7x23x29=2x11x17+3x13x19x31

496961=1235+495726  17x23x31x41=5x13x19+2x3x7x11x29x37
219965=6479+213486  5x29x37x41=11x19x31+2x3x7x13x17x23
144210=16523+127687 2x3x5x11x19x23=13x31x41+7x17x29x37
226083=6118+219965  3x11x13x17x31=2x7x19x23+5x29x37x41
106981=57646+49335  7x17x29x31=2x19x37x41+3x5x11x13x23

No further solutions with n<=35

Q3:

I computed the minimum absolute distance between b^2+4P/b and the nearest square (where no solution was found), let's call this dist(D^2). The smallest distance 103 was found (n in [12,14-26]) at n=18, b=12509182. At first sight this is large, however a and c fail to be integer by a mere 0.000000133 (=dist(a),dist(b)). However the resulting distance between P and a.b.c is huge, since a.b.c=(D+b)/2.b.(D-b)/2=b.(D^2-b^2)/4, this gap is of the order b.dist(D^2)/4=12509182.103/4.

In general dist(a)=dist(b)=dist(D^2)/(4D) and dist(P)=b.dist(D^2)/4. And dist(D^2) is maximal D+1/4. More should be known about the distribution of dist(D^2) to say more.

***

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 Home | Melancholia | Problems & Puzzles | References | News | Personal Page | Puzzlers | Search | Bulletin Board | Chat | Random Link Copyright © 1999-2012 primepuzzles.net. All rights reserved.
user
The text passage contains a series of mathematical puzzles and questions, along with some provided solutions and additional commentary from contributors.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS

1. The passage contains multiple mathematical questions (Questions 1, 2, and 3).
2. Answers to these questions are provided in the form of examples and solutions contributed by Hakan Summakoğlu and Jan van Delden.