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Some ETs follow a positional number system, with the same base as the number of fingers on their hand. The following inscription is all the evidence we have: $$(\Box @)+(\Box @) = \Box\bigstar\Box $$ Find the number of fingers.

Attempt: $(xy)_b + (xy)_b = (xzx)_b$ Can be written as $2y+2bx=x+bz+b^2 x$ so I need to solve for $b$ the equation $b^2x+bz-2bx+x-y=0$ with $b\geq 2$ and $0\leq x,y,z<b $.

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Regarding Source, M Gardner's mathematical games was mentioned in the book, but I havent been able to locate it. Please post an online link if you know it. –  kuch nahi Jun 26 '11 at 9:41
my interpretation of $(\Box @)+(\Box @) = (\Box\bigstar\Box)$ is $(xy)_b+(xy)_b=(xzx)_b$ and therefore $(bx+y)+(bx+y)=b^2x+zb+x$, which can be transformed into $2y+2xb=(b^2+1)x+bz$ –  miracle173 Jun 26 '11 at 9:42
Hint: When adding two double 'digit' integers together, the maximum carry you can get to the third position is equal to ... –  Jyrki Lahtonen Jun 26 '11 at 9:43
@miracle173 made the correction –  kuch nahi Jun 26 '11 at 9:45
@Jyrki ..... 1? –  kuch nahi Jun 26 '11 at 9:53

2 Answers 2

up vote 22 down vote accepted

You have tried to solve it more algebraically than normal people would do. (I usually do the same thing - too scientific an approach - which occasionally makes us less efficient in similar puzzles.)

Spoilers are found below. Please stop reading if you want to solve it yourself.

Each of the two 2-digit expressions are smaller than $b^2$, so their sum is smaller than $2b^2$. Because this is a 3-digit expression, its first digit has to be $1$. So the square is equal to 1. So we have $$(1@) + (1@) = (1\bigstar 1)$$ Note that it ends with a $1$ - odd digit - even though the left-hand side is even - twice $(1@)$. It can only happen if the base $b$ is odd.

At any rate, the equation simplifies to $$2(b+@) = b^2+1 + b\bigstar$$ Moving everything non-negative to the right hand side, we have $$ 2@ = (b-1)^2 + b \bigstar.$$ The first term is a square of an integer, i.e. $4,16,36,64,\dots $ for $b=3,5,7,9,\dots$. Note that the identity above implies that $2@$ can't be smaller than $(b-1)^2$ but $2@<2b$ and if $b\geq 5$, $2b$ is clearly smaller than $(b-1)^2$. So the only chance to find a solution is $b=3$. Then the equation simplifies to $$ 2@ = 4+3 \bigstar .$$ The left hand side is even, so the right hand side must also be even. Therefore $\bigstar$ is either $0$ or $2$. For $\bigstar=2$ we would get $@=5$ which is too high so the only other option is $\bigstar=0$ and $@=2$ which works: $$ (12) + (12) = (101).$$ This trinary equation is translated to decimal base as $5+5=10$. Note that those ETs have 3 fingers in total, so if they have an even number of hands, it follows that different hands have different numbers of fingers, for example 1 left finger and 2 right fingers.

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I'm curious as to why this was downvoted. –  barf Jun 26 '11 at 11:39
Me too. I posted about ten minutes earlier, but that's hardly a reason to downvote. –  Jyrki Lahtonen Jun 26 '11 at 18:04
Thanks for your care and implicit compliments, @JTL and @Jyrki, but I don't think it's a big issue. Let's be generous, people have the right to vote the way they want. –  Luboš Motl Jun 27 '11 at 20:08
@barf: it was probably downvoted because there's an earlier answer that seems complete enough. –  Zsbán Ambrus Mar 4 '12 at 12:08

I would rather hide this behind a spoiler so that others can have fun, too. If there is a way of doing it, @-message me, please.

The only carry to the third postion that we can get in the sum of two double digit numbers is 1. Therefore `square'='1'. The sum is obviously an even number, so as the sum ends with a '1' = odd digit, the base 'b' must be an odd number. The double digit integer $x=$ 'square @' is $<2b$. Because of the carry we get that $2x\ge b^2$, so $4b>b^2$, and therefore $b<4$. The only alternative for the base is thus $b=3$. $y=$'@' must satisfy $2y>b$, because otherwise the least significant digit couldn't overflow, so '@' must be $=2$. The only remaining possibility is $$ 12_3+12_3=101_3, $$ which checks out.

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there has been a very recent discussion of spoilers on meta. –  Gerry Myerson Jun 26 '11 at 10:58
@Gerry: Thanks. I will search for more of that discussion later. I was looking for a hide/unhide button and that seems to be on the wish list, too. They have one at XKCD-fora, and it works well for cases like this, where a 'teacher' really would like to encourage others to work it out themselves, but can't stick around to discuss the problem by giving a sequence of hints as comments. –  Jyrki Lahtonen Jun 26 '11 at 11:07
1… shows you how to have a gray box that text appears in when moused over. You lead with greater than exclamation point. –  Ross Millikan Jun 26 '11 at 16:18
The software is really not geared towards the type of questions where a hide/unhide button should be necessary. If a given person doesn't want to read an answer, they simply shouldn't scroll down. –  Qiaochu Yuan Jun 26 '11 at 16:25

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