HMMT 二月 2003 · 代数 · 第 3 题

HMMT February 2003 — Algebra — Problem 3

解析

Find the smallest n such that n ! ends in 290 zeroes. Solution: 1170 Each 0 represents a factor of 10 = 2 · 5. Thus, we wish to find the smallest factorial that contains at least 290 2’s and 290 5’s in its prime factorization. Let this number be n !, so the factorization of n ! contains 2 to the power p and 5 to the power q , where ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ n n n n n n p = + + + · · · and q = + + + · · · 2 3 2 3 2 2 2 5 5 5 (this takes into account one factor for each single multiple of 2 or 5 that is ≤ n , 2 2 an additional factor for each multiple of 2 or 5 , and so on). Naturally, p ≥ q because 2 is smaller than 5. Thus, we want to bring q as low to 290 as possible. n n n If q = b c + b c + b c + · · · , we form a rough geometric sequence (by taking away 2 3 5 5 5 n/ 5 the floor function) whose sum is represented by 290 ≈ . Hence we estimate 1 − 1 / 5 n = 1160, and this gives us q = 288. Adding 10 to the value of n gives the necessary two additional factors of 5, and so the answer is 1170. √ √ √