HMMT 十一月 2012 · 冲刺赛 · 第 11 题

HMMT November 2012 — Guts Round — Problem 11

[ 8 ] Find the smallest positive integer n such that the number of zeroes that n ! ends with is a positive multiple of 5.

解析

[ 8 ] Answer: 45 The number of zeroes that n ! ends with is the largest power of 10 dividing n !. The exponent of 5 dividing n ! exceeds the exponent of 2 dividing n !, so we simply seek the exponent of 5 dividing n !. For a number less than 125, this exponent is just the number of multiples of 5, but not 25, less than n plus twice the number of multiples of 25 less than n . Counting up, we see that 24! ends with 4 zeroes while 25! ends with 6 zeroes, so n ! cannot end with 5 zeroes. Continuing to count up, we see that the smallest n such that n ! ends with 10 zeroes is 45.