PUMaC 2017 · 团队赛 · 第 15 题

PUMaC 2017 — Team Round — Problem 15

解析

Given yx = y , we can rewrite this to obtain x = y = ⇒ x = y . Therefore, we 2016 y want to find positive integer values of y that make the quantity y an integer. We can split 2016 the potential solutions for y into 2 sets: where is an integer, and where it is not. The first y 5 2 set is simple; the prime factorization of 2016 is 2 · 3 · 5, so there are 6 · 3 · 2 = 36 factors of 2016, each of which generates a unique, valid solution ( x, y ). The second set can be computed as follows: let y = kd , where d is a (maximal) divisor of 2016 and y is not a divisor of 2016. 2016 1 2016 kd k d Then, ( kd ) = (( kd ) ) . A sufficient condition for this quantity to be an integer is that 1 th k kd is a perfect k power, so that ( kd ) can be an integer. 5 • If k = 2, and since kd is not a factor of 2016, we need d = 2 · n , where n is a perfect 2 square factor, namely either 1 or 3 , yielding 2 solutions. 2 3 • If k = 3, we need d = 3 · n , where n is a perfect cube factor, namely either 1 or 2 , yielding 2 solutions. • If k = 4, 5, 6, or 7, there are no solutions. 5 3 th 5 • If k = 8, we need 2 · n to be a perfect 8 power, namely n = 2 is the only possible, yielding 1 more. This yields a total of 36 + 5 = 41 solutions. Problem written by Jackson Blitz; solution written by Juneseo Lee ∑ ∑