# 1962 AHSME Problems/Problem 38

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## Problem

The population of Nosuch Junction at one time was a perfect square. Later, with an increase of $100$, the population was one more than a perfect square. Now, with an additional increase of $100$, the population is again a perfect square.

The original population is a multiple of:

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 17$

## Solution

Let $a^2$ $=$ original population count, $b^2+1$ $=$ the second population count, and $c^2$ $=$ the third population count We first see that $a^2 + 100 = b^2 + 1$ or $99$ $=$ $b^2-a^2$. We then factor the right side getting $99$ $=$ $(b-a)(b+a)$. Since we can only have an nonnegative integral population, clearly $b+a$ $>$ $b-a$ and both factor $99$. We factor $99$ into $3^2 \cdot 11$ $=$ $(b-a)(b+a)$ There are a few cases to look at: $1)$ $b+a$ $=$ $11$ and $b-a$ $=$ $9$. Adding the two equations we get $2b$ $=$ $20$ or $b$ $=$ $10$, which means $a$ $=$ $1$. But looking at the restriction that the second population + $100$ $=$ third population... $10^2$ $+$ $1$ $+$ $100$ $=$ $201$ $\neq$ a perfect square.

$2)$ $b+a$ $=$ $33$ and $b-a$ $=$ $3$. Adding the two equations we get $2b$ $=$ $36$ or $b$ $=$ $18$, which means $a$ $=$ $15$. Looking at the same restriction, we get $18^2$ + $1$ + $100$ $=$ $324$ + $101$ $=$ $425$, which is NOT a perfect square.

Finally, $b+a$ $=$ $99$ and $b-a$ $=$ $1$. $2b$ $=$ $100$ or $b$ $=$ $50$, which means $a$ $=$ $49$. Looking at the same restriction, we get $50^2$ + $1$ + $100$ $=$ $2500$ + $101$ $=$ $2601$ $=$ $51^2$. Thus we find that the original population is $a^2$ $=$ $49^2$ $=$ $7^4$. Or $a^2$ is a multiple of $\boxed{ (B) 7}$