Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $.

share|improve this question
    
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.

share|improve this answer
3  
I'm curious as to why this was downvoted. –  barf Jun 26 '11 at 11:39
2  
Me too. I posted about ten minutes earlier, but that's hardly a reason to downvote. –  Jyrki Lahtonen Jun 26 '11 at 18:04
6  
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.

share|improve this answer
    
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  
meta.stackexchange.com/questions/1191/… 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
4  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.