HMMT 二月 2016 · 代数 · 第 4 题

HMMT February 2016 — Algebra — Problem 4

Determine the remainder when ⌊ ⌋ 2015 i ∑ 2 25 i =0 is divided by 100, where b x c denotes the largest integer not greater than x .

解析

Determine the remainder when ⌊ ⌋ 2015 i ∑ 2 25 i =0 is divided by 100, where b x c denotes the largest integer not greater than x . Proposed by: Alexander Katz Answer: 14 i φ (25) 20 Let r denote the remainder when 2 is divided by 25. Note that because 2 ≡ 2 ≡ 1 (mod 25), i i r is periodic with length 20. In addition, we find that 20 is the order of 2 mod 25. Since 2 is never a multiple of 5, all possible integers from 1 to 24 are represented by r , r , ..., r with the exceptions of 1 2 20 ∑ ∑ 20 24 5, 10 ,15, and 20. Hence, r = i − (5 + 10 + 15 + 20) = 250. i i =1 i =1 We also have ⌊ ⌋ 2015 2015 i i ∑ ∑ 2 2 − r i = 25 25 i =0 i =0 2015 2015 i ∑ ∑ 2 r i = − 25 25 i =0 i =0 1999 15 2016 ∑ ∑ 2 − 1 r r i i = − − 25 25 25 i =0 i =0 ( ) 15 2016 ∑ 2 − 1 250 r i = − 100 − 25 25 25 i =0 15 2016 ∑ 2 − 1 r i ≡ − (mod 100) 25 25 i =0 ∑ 15 We can calculate r = 185, so i i =0 ⌊ ⌋ 2015 i 2016 ∑ 2 2 − 186 ≡ (mod 100) 25 25 i =0 φ (625) 500 2016 16 2016 Now 2 ≡ 2 ≡ 1 (mod 625), so 2 ≡ 2 ≡ 536 (mod 625). Hence 2 − 186 ≡ 350 2016 2016 (mod 625), and 2 − 186 ≡ 2 (mod 4). This implies that 2 − 186 ≡ 350 (mod 2500), and so 2016 2 − 186 ≡ 14 (mod 100). 25