HMMT 二月 2025 · 冲刺赛 · 第 11 题

HMMT February 2025 — Guts Round — Problem 11

[7] Let f ( n ) = n + 100. Compute the remainder when f ( f ( · · · f ( f ( 1)) · · · )) is divided by 10 . | {z } 2025 f ’s

解析

[7] Let f ( n ) = n + 100. Compute the remainder when f ( f ( · · · f ( f ( 1)) · · · )) is divided by 10 . | {z } 2025 f ’s Proposed by: Pitchayut Saengrungkongka Answer: 3101 k n 4 Solution: We claim that f ( n ) ≡ 1 + 100(2 − 1) mod 10 . We can see this by induction, as n n 2 f (1 + 100(2 − 1)) = (1 + 100(2 − 1)) + 100 n 4 ≡ 1 + 200(2 − 1) + 100 (mod 10 ) n +1 4 ≡ 1 + 100(2 − 1) (mod 10 ) . 2025 2025 Thus, it suffices to compute 2 mod 100. We note that 2 ≡ 0 (mod 4) and by Euler’s Totient 2025 5 2025 theorem, 2 ≡ 2 (mod 25), so 2 ≡ 32 (mod 100). Hence, we can compute 2025 4 f (1) ≡ 1 + 100(31) ≡ 3101 mod 10 .