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2025/26 AHSMC Part 1 Problem 7

Problem 7 of 16

How many positive integers nn satisfy n2−15n=51−35⋅(2+(−1)n)? n^2 - 15n = 51 - 35\cdot\bigl(2 + (-1)^n\bigr)?

Answer choices

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Solution

If nn is odd the equation is n2−15n=16n^2 - 15n = 16, with solutions n=−1n = -1 and n=16n = 16, neither a positive odd integer. If nn is even the equation is n2−15n=−54n^2 - 15n = -54, with solutions n=6n = 6 and n=9n = 9; only n=6n = 6 is even. There is one such nn. The answer is (B).

University of Alberta