HMMT 二月 2003 · 冲刺赛 · 第 15 题

HMMT February 2003 — Guts Round — Problem 15

[7] The product of the digits of a 5-digit number is 180. How many such numbers exist? 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

解析

The product of the digits of a 5-digit number is 180. How many such numbers exist? Solution: 360 2 2 Let the digits be a, b, c, d, e. Then abcde = 180 = 2 · 3 · 5 . We observe that there are 6 ways to factor 180 into digits a, b, c, d, e (ignoring differences in ordering): 180 = 1 · 1 · 4 · 5 · 9 = 1 · 1 · 5 · 6 · 6 = 1 · 2 · 2 · 5 · 9 = 1 · 2 · 3 · 5 · 6 = 1 · 3 · 3 · 4 · 5 = 2 · 2 · 3 · 3 · 5 . There are (respectively) 60, 30, 60, 120, 60, and 30 permutations of these breakdowns, for a total of 360 numbers. 1