# 1967 AHSME Problems/Problem 36

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## Problem

Given a geometric progression of five terms, each a positive integer less than $100$. The sum of the five terms is $211$. If $S$ is the sum of those terms in the progression which are squares of integers, then $S$ is: $\textbf{(A)}\ 0\qquad \textbf{(B)}\ 91\qquad \textbf{(C)}\ 133\qquad \textbf{(D)}\ 195\qquad \textbf{(E)}\ 211$

## Solution

Let the first term be $a$ and the common ratio be $r$, and WLOG let $r \ge 1$. The five terms are $a, ar, ar^2, ar^3, ar^4$, and the sum is $a(1 + r + r^2 + r^3 + r^4)$. Clearly $r$ must be rational for all terms to be integers. If $r$ were an integer, it could not be $1$, since $a$ would equal $\frac{211}{1 + r + r^2 + r^3 + r^4}$, which is not an integer. In fact, quickly testing $r = 2, 3, 4$ shows that $r$ cannot be an integer.

We now consider non-integers. If $r = \frac{x}{y}$ and $\gcd(x, y) = 1$, then $a$ would have to be divisible by $y^4$, since $ar^4 = a\frac{x^4}{y^4}$ is an integer. If $y = 3$, then $a$ would have to be a multiple of $81$, which would make the five terms sum to at least $5 \cdot 81$. It only gets worse if $y > 3$. Thus, $y = 2$, and $a$ is a multiple of $2^4 = 16$. Let $a = 16k$.

We now look at the last term, $ar^4 = 16k\frac{x^4}{16} = kx^4$. The smallest allowable values for $x$, given that $x$ cannot be even, are $3$ and $5$. If $x = 5$, the last term will be way too big. Thus, $x = 3$ is the only possibility.

We now have a sum of $a(1 + r + r^2 + r^3 + r^4) = 211$, and we know that $r = \frac{3}{2}$ is the only possibility. The terms in the parentheses happen to equal $\frac{211}{16}$ when you plug them in, so $a = 16$.

Thus, the terms are $16, 24, 36, 54, 81$, and the first, third, and fifth terms are squares, with a sum of $133$, which is answer $\fbox{C}$.

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