HMMT 二月 2026 · ALGNT 赛 · 第 2 题

HMMT February 2026 — ALGNT Round — Problem 2

Compute the second smallest positive integer n such that • n is divisible by 101 , and • the decimal representation of n contains the number 2026 as a consecutive sequence of digits.

解析

Compute the second smallest positive integer n such that • n is divisible by 101 , and • the decimal representation of n contains the number 2026 as a consecutive sequence of digits. Proposed by: Jacopo Rizzo Answer: 420261 Solution: We proceed by casework on the position of the substring 2026 in n . • If n ends in 2026 , then we may write n = 10000 x + 2026 for some positive integer x . Then, the condition 101 | n simplifies to 101 | x + 6 . Hence, x ≥ 95 , so the smallest possibility in this case is 952026 . • If n ends in 2026 followed by a digit, then we may write n = 100000 x + 20260 + y , where x is a positive integer and y is a digit. The condition 101 | n simplifies to 101 | 10 x + 60 + y . Hence, 10 x + y ≥ 41 , so the smallest possibility in this case is 420261 . • If n ends in 2026 followed by two digits, then we may write n = 1000000 x + 202600 + y , where x is a positive integer and y is a two-digit positive integer. The condition 101 | n simplifies to 101 | ( − x − 6 + y ) . If x = 0 , then the only possibility is y = 6 , yielding the solution 202606 . 6 Otherwise, x > 0 , yielding solutions n ≥ 10 . 6 • If n ends in 2026 followed by three or more digits, then n ≥ 10 . Thus, the smallest two values of n are 202606 and 420261 .