HMMT 二月 2001 · 代数 · 第 3 题

HMMT February 2001 — Algebra — Problem 3

解析

How many times does 24 divide into 100! (factorial)? Solution: We first determine the number of times 2 and 3 divide into 100! = 1 · 2 · 3 · · · 100. Let 〈 N 〉 be the number of times n divides into N (i.e. we want to find 〈 100! 〉 ). Since 2 n 24 only divides into even integers, 〈 100! 〉 = 〈 2 · 4 · 6 · · · 100 〉 . Factoring out 2 once from each 2 50 of these multiples, we get that 〈 100! 〉 = 〈 2 · 1 · 2 · 3 · · · 50 〉 . Repeating this process, we 2 2 50+25+12+6+3+1 33+11+3+1 find that 〈 100! 〉 = 〈 20 · 1 〉 = 97. Similarly, 〈 100! 〉 = 〈 3 〉 = 48. 2 2 3 3 3 Now 24 = 2 · 3, so for each factor of 24 in 100! there needs to be three multiples of 2 and one multiple of 3 in 100!. Thus 〈 100! 〉 = ([ 〈 100! 〉 / 3] + 〈 100! 〉 ) = 32 , where [ N ] is the 24 2 3 greatest integer less than or equal to N .