HMMT 二月 2002 · 代数 · 第 6 题
HMMT February 2002 — Algebra — Problem 6
题目详情
- Find the sum of the even positive divisors of 1000.
解析
- Find the sum of the even positive divisors of 1000. 1 Solution: 2184 . Notice that 2 k is a divisor of 1000 iff k is a divisor of 500, so we need only find the sum of the divisors of 500 and multiply by 2. This can be done by enumerating 2 3 2 the divisors individually, or simply by using the formula: σ (2 · 5 ) = (1 + 2 + 2 )(1 + 5 + 2 3 5 + 5 ) = 1092, and then doubling gives 2184. Alternate Solution: The sum of all the 2 3 2 3 divisors of 1000 is (1 + 2 + 2 + 2 )(1 + 5 + 5 + 5 ) = 2340. The odd divisors of 1000 are 2 3 simply the divisors of 125, whose sum is 1 + 5 + 5 + 5 = 156; subtracting this from 2340, we are left with the sum of the even divisors of 1000, which is 2184.