HMMT 十一月 2012 · 冲刺赛 · 第 11 题
HMMT November 2012 — Guts Round — Problem 11
题目详情
英文原题
- [ 8 ] Find the smallest positive integer n such that the number of zeroes that n ! ends with is a positivemultiple of 5.
解析
英文解析
- [ 8 ]
Answer: 45 The number of zeroes that n ! ends with is the largest power of 10 dividing n !. Theexponent 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 nplus 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.