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We live in a community of houses sequentially numbered from 1 to 100. We all love square dancing but only two immediate neighbors are joy to watch. If you concatenate their house numbers, it forms a true square.

Who are they?

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  • $\begingroup$ Very nice puzzle! I'll admit that I found this by checking a lot of possibilities (by hand, not programming anything), but there's a very neat mathematical solution which yields the solution almost without any calculation at all. $\endgroup$ – Rand al'Thor May 9 at 18:58
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    $\begingroup$ Thx..during my long walks I come up with these..I love to do mental math..I can square most of the numbers upto 1000 mentally.. $\endgroup$ – Uvc May 9 at 19:05
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The answer is

$8281$, the concatenation of 82 and 81, which is $91^2$.

Proof:

We're looking for a perfect square of the form $xyxz$ where $z=y\pm1$. But this equals $101(10x+y)\pm1$, which is congruent to $\pm1$ modulo $101$. So its square root must be less than $100$ but congruent to a square root of $\pm1$ modulo 101.

One of the two cases can be eliminated immediately:

The square roots of $+1$ modulo $101$ are $\pm1$, which are not possibilities. (Note that $101$ is prime, so there are exactly two square roots.)

So we check the other one.

What are the square roots of $-1$ modulo $101$? Clearly $10^2\equiv-1$, but that doesn't give us the solution we need. The other square root is $-10\equiv91$, and that does give us the solution.

QED.

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  • $\begingroup$ Very neat mathematical proof for general case $\endgroup$ – Uvc May 10 at 2:15

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