HMMT 十一月 2013 · 冲刺赛 · 第 18 题

HMMT November 2013 — Guts Round — Problem 18

[ 10 ] The rightmost nonzero digit in the decimal expansion of 101! is the same as the rightmost nonzero digit of n !, where n is an integer greater than 101. Find the smallest possible value of n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT NOVEMBER 2013, 9 NOVEMBER 2013 — GUTS ROUND

解析

[ 10 ] The rightmost nonzero digit in the decimal expansion of 101! is the same as the rightmost nonzero digit of n !, where n is an integer greater than 101. Find the smallest possible value of n . Answer: 103 101! has more factors of 2 than 5, so its rightmost nonzero digit is one of 2 , 4 , 6 , 8. Notice that if the rightmost nonzero digit of 101! is 2 k (1 ≤ k ≤ 4), then 102! has rightmost nonzero digit 102(2 k ) ≡ 4 k (mod 10), and 103! has rightmost nonzero digit 103(4 k ) ≡ 2 k (mod 10). Hence n = 103.