HMMT 二月 2003 · 代数 · 第 3 题

英文解析

3. 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 factorialthat contains at least 290 2’s and 290 5’s in its prime factorization. Let this numberbe n !, so the factorization of n ! contains 2 to the power p and 5 to the power q , where
   ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ ⌊ ⌋
   n n n n n np = + + + · · · 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 ≥ qbecause 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 necessarytwo additional factors of 5, and so the answer is 1170.
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