1967 AHSME Problems/Problem 36
Problem
Given a geometric progression of five terms, each a positive integer less than 
. The sum of the five terms is 
. If 
is the sum of those terms in the progression which are squares of integers, then is:
Solution
Let the first term be 
and the common ratio be 
, and WLOG let 
. The five terms are 
, and the sum is 
. Clearly must be rational for all terms to be integers. If were an integer, it could not be 
, since would equal 
, which is not an integer. In fact, quickly testing 
shows that cannot be an integer.
We now consider non-integers. If 
and 
, then would have to be divisible by 
, since 
is an integer. If 
, then would have to be a multiple of 
, which would make the five terms sum to at least 
. It only gets worse if 
. Thus, 
, and is a multiple of 
. Let 
.
We now look at the last term, 
. The smallest allowable values for 
, given that cannot be even, are 
and 
. If 
, the last term will be way too big. Thus, 
is the only possibility.
We now have a sum of 
, and we know that 
is the only possibility. The terms in the parentheses happen to equal 
when you plug them in, so 
.
Thus, the terms are 
, and the first, third, and fifth terms are squares, with a sum of 
, which is answer 
.
See also
| 1967 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 35 |
Followed by Problem 37 | |
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