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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