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1962 AHSME Problems/Problem 38

Revision as of 15:52, 28 June 2023 by Nullptr07 (talk | contribs) (→‎Solution)
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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 1

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 $\boxed{ (B) 7}$

Solution 2

Let $P$ $=$ original population. Translating the word problem into a system of equations, we got: \begin{align} P &= x^2 \\ P + 100 &= y^2 + 1 \\ P + 200 &= z^2 \end{align} for some positive integers $x$, $y$ and $z$. Now, by subtracting $(2)$ from $(3)$ (i.e. $(3) - (2)$), we got: \begin{align*} 100 &= z^2 - y^2 - 1 \\ 101 &= z^2 - y^2 \\ 101 &= (z - y)(z + y) \end{align*} Since y and z are both positive integers and 101 is a prime, by factoring, the only working solution for us is $y = 50$ and $z = 51$. Plugging that back to $(2)$ and simplify, we got $P = 2401 = (49)^2 = x^2$, a multiple of $7$. Therefore, the answer is $\boxed{(B) 7}$. -nullptr07

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