HMMT 十一月 2021 · 团队赛 · 第 4 题

HMMT November 2021 — Team Round — Problem 4

[35] Find the number of 10-digit numbers a a · · · a which are multiples of 11 such that the digits 1 2 10 are non-increasing from left to right, i.e. a ≥ a for each 1 ≤ i ≤ 9. i i +1

解析

[35] Find the number of 10-digit numbers a a · · · a which are multiples of 11 such that the digits 1 2 10 are non-increasing from left to right, i.e. a ≥ a for each 1 ≤ i ≤ 9. i i +1 Proposed by: Sheldon Kieren Tan Answer: 2001 Solution: It is well known that a a · · · a is divisible by 11 if and only if S = a − a + a − · · · − a 1 2 10 1 2 3 10 is. By the non-increasing condition, we deduce that S = ( a − a ) + ( a − a ) + · · · + ( a − a ) ≥ 0 . 1 2 3 4 9 10 Also, S = a − ( a − a ) − · · · − ( a − a ) − a ≤ a ≤ 9 . 1 2 3 8 9 10 1 Therefore, S = 0, our number must be of the form aabbccddee . Since all numbers of this form work ( aabbccddee = 11 · a 0 b 0 c 0 d 0 e ), it suffices to find the number of tuples ( a, b, c, d, e ) so that 9 ≥ a ≥ b ≥ c ≥ d ≥ e ≥ 0 and a > 0. The number of tuples satisfying the first condition is ( ) 14 = 14 · 13 · 11 = 2002. To account for the second condition, we subtract one tuple (all zeroes), 5 yielding a final answer of 2001.