PUMaC 2014 · 数论（B 组） · 第 3 题

PUMaC 2014 — Number Theory (Division B) — Problem 3

解析

[ 4 ] Find the 3-digit positive integer that has the most divisors. Solution: Since 2 × 3 × 5 × 7 × 11 > 1000, the most number of different prime divisors a 3-digit positive integer can have is 4, and they should be the smallest one. 10 If the integer has only 1 prime divisor, then since 2 > 1000, it can have at most 10 divisors. 6 3 If the integer has 2 prime divisors, then 2 × 3 > 1000, so the integer has at most 6 × 4 = 24 divisors. 5 2 If the integer has 3 prime divisors, then 2 × 3 × 5 > 1000, so the integer has at most 5 × 3 × 2 = 30 divisors. 4 If the integer has 4 prime divisors, then 2 × 3 × 5 × 7 > 1000, so the integer has at most 4 × 2 × 2 × 2 = 32 divisors. 3 Thus, 2 × 3 × 5 × 7 = 840 has the most divisors. p