PUMaC 2022 · 数论（B 组） · 第 4 题

PUMaC 2022 — Number Theory (Division B) — Problem 4

解析

Compute the sum of all positive integers whose positive divisors sum to 186. Proposed by Nancy Xu Answer: 202 n n n 1 2 k The sum of the divisors of an integer with prime factorization p p . . . p is given by (1 + 1 2 k n n n 1 2 k p + . . . p )(1 + p + . . . p ) . . . (1 + p + . . . p ). We see that 186 = 2 · 3 · 31, so it has 1 2 k 1 2 k factors 1 , 2 , 3 , 6 , 31 , 62 , 93 , 186. It is clear that 1 and 2 cannot be written as the sum of powers of a prime, so by trying out small primes, the only remaining possibilities are 186 = 6 · 31 = (1 + 5)(1 + 2 + 4 + 8 + 16) and 186 = 3 · 62 = (1 + 2)(1 + 61). Thus our two numbers are 5 · 16 = 80 and 2 · 61 = 122, sum the sum is 80 + 122 = 202.