PUMaC 2011 · 数论（B 组） · 第 3 题

PUMaC 2011 — Number Theory (Division B) — Problem 3

[ 4 ] The only prime factors of an integer n are 2 and 3. If the sum of the divisors of n (including itself) is 1815, find n .

解析

The sum of the divisors of n = 2 3 is equal to (1+2 +2 + · · · +2 )(1+3 +3 + · · · +3 ) = 1815 , i j since each divisor of 2 3 is represented exactly once in the sum that results when the product 1 2 i i +1 1 2 j is expanded. Let A = 1 + 2 + 2 + · · · + 2 = 2 − 1 and B = 1 + 3 + 3 + · · · + 3 , so that 2 AB = 1815 = 3 · 5 · 11 . i +1 Since B ≡ 1 (mod 3), 3 | A . By Fermat’s Little Theorem, 2 − 1 ≡ 0 (mod 3) only when i is odd. For i = 1 we get A = 3 , B = 605 which does not work. For i = 3 we get A = 15 , B = 121, which holds for n = 648 . For i = 5 , 7 , 9 , 7 | A, 17 | A, 31 | A, respectively, and for i > 10 , A > 1815.