PUMaC 2022 · 数论（B 组） · 第 1 题

PUMaC 2022 — Number Theory (Division B) — Problem 1

Suppose that the greatest common divisor of n and 5040 is equal to 120 . Determine the sum of the four smallest possible positive integers n.

解析

Suppose that the greatest common divisor of n and 5040 is equal to 120 . Determine the sum of the four smallest possible positive integers n. Proposed by Frank Lu Answer: 3600 Note that for the greatest common divisor of n and 5040 to equal 120 , we need n = 120 d, where 5040 d is relatively prime to = 42 . But then note that this means that d can’t be divisible by 120 2 , 3 , or 7 . This yields us that d = 1 , 5 , 11 , 13 , yielding the sum of n as 120(1 + 5 + 11 + 13) = 120(30) = 3600 .