HMMT 二月 2026 · 冲刺赛 · 第 13 题
HMMT February 2026 — Guts Round — Problem 13
题目详情
- [9] The concatenation of two base- 10 numbers (possibly with leading 0 s) is defined as the base- 10 number formed by joining them together. For example, the concatenation of 1402 and 00213 is 140200213 . Compute the number of 2026 -digit multiples of 3 which cannot be expressed as the concatenation of two smaller multiples of 3 (possibly with leading 0 s).
解析
- [9] The concatenation of two base- 10 numbers (possibly with leading 0 s) is defined as the base- 10 num- ber formed by joining them together. For example, the concatenation of 1402 and 00213 is 140200213 . Compute the number of 2026 -digit multiples of 3 which cannot be expressed as the concatenation of two smaller multiples of 3 (possibly with leading 0 s). Proposed by: Justin Zhang, Sidarth Erat 2024 Answer: 18 · 7 Solution: Let n be a 2026 -digit multiple of 3 . Then n satisfies the desired condition if and only if for all m from 1 to 2025 , the first m digits of n have sum not divisible by 3 . To compute the number of such n , we will construct n one digit at a time. ©2026 HMMT • There are 6 ways to pick the first digit, as it can be any non-multiple of 3 . (any digits except 0 , 3 , 6 , 9 ) • For each m = 2 , 3 , . . . , 2025 , given that the sum of the first m − 1 digits of n , called d , is not divisible by 3 . th – If d ≡ 1 (mod 3) . Then, the m digit of n can neither be 2 , 5 , nor 8 , so there are only 7 remaining choices. th – If d ≡ 2 (mod 3) . Then, the m digit of n can neither be 1 , 4 , nor 7 , so there are only 7 remaining choices. th In either case, there are only 7 ways to pick the m digit. • To pick the last digit, not that the sum of all digits must be divisible by 3 . If the sum of the first 2025 digits has remainder 1 when divided by 3 , the last digit can only be 2 , 5 , 8 . Similarly, if the remainder is 2 instead, the last digit can only be 1 , 4 , 7 . In either case, there are 3 ways to choose the last digit. Hence, the answer is 2024 2024 6 · 7 · 3 = 18 · 7 .