1989 AHSME Problems/Problem 27
Contents
Problem
Let be a positive integer. If the equation has 28 solutions in positive integers , , and , then must be either
Solution
This is equivalent to finding the powers of with coefficient in the expansion of .
But this is
, the last part having general term from which it is easy to see that, since , the last part contains the term and the whole result . So the answer is .
Solution 2
Suppose is even. Thus is even, so let . Hence . By stars and bars, this has solutions. This implies that . If is odd, let . Then , in which is a solution. The answer is then .
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See also
1989 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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