HMMT 二月 2025 · 冲刺赛 · 第 13 题

HMMT February 2025 — Guts Round — Problem 13

[9] A number is upwards if its digits in base 10 are nondecreasing when read from left to right. Compute 6 the number of positive integers less than 10 that are both upwards and multiples of 11.

解析

[9] A number is upwards if its digits in base 10 are nondecreasing when read from left to right. Compute 6 the number of positive integers less than 10 that are both upwards and multiples of 11. Proposed by: Srinivas Arun Answer: 219 Solution: For a number d d d d d d (allowing leading 0s) to be upwards and a multiple of 11, we 5 4 3 2 1 0 must have d ≤ d ≤ d ≤ d ≤ d ≤ d , 5 4 3 2 1 0 d − d + d − d + d − d ≡ 0 (mod 11) . 0 1 2 3 4 5 Note that d − d , d − d , and d − d are all nonnegative. Thus, 0 1 2 3 4 5 0 ≤ ( d − d ) + ( d − d ) + ( d − d ) 0 1 2 3 4 5 ≤ ( d − d ) + ( d − d ) + ( d − d ) + ( d − d ) + ( d − d ) 0 1 1 2 2 3 3 4 4 5 = d − d 0 5 ≤ 9 . Therefore, ( d − d ) + ( d − d ) + ( d − d ) = 0 , 0 1 2 3 4 5 which can only occur when d = d , d = d , and d = d , i.e. the number is of the form aabbcc . We 0 1 2 3 4 5 can easily verify that all numbers of the form aabbcc for digits a ≤ b ≤ c satisfy our conditions, so we simply have to count them. 12 There are = 220 such triples of digits ( a, b, c ). However, one of these triples is (0 , 0 , 0), which 3 corresponds to the number 0. Thus our answer is 220 − 1 = 219 .