HMMT 二月 2003 · 冲刺赛 · 第 18 题
HMMT February 2003 — Guts Round — Problem 18
题目详情
- [8] Find the sum of the reciprocals of all the (positive) divisors of 144. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND 3 2
解析
- Find the sum of the reciprocals of all the (positive) divisors of 144. Solution: 403 / 144 As d ranges over the divisors of 144, so does 144 /d , so the sum of 1 /d is 1 / 144 times the sum of the divisors of 144. Using the formula for the sum of the divisors of a number (or just counting them out by hand), we get that this sum is 403, so the answer is 403 / 144. 4 3 2