PUMaC 2023 · 代数（B 组） · 第 7 题
PUMaC 2023 — Algebra (Division B) — Problem 7
题目详情
- Compute 3 + 2 cos (mod 100). 11 k =0 P 2 n − 1 j
解析
- Compute 3 + 2 cos (mod 100). 11 k =0 Proposed by Sunay Joshi and Ben Zenker Answer: 91 Let n = 10. We claim that the sum equals ⌊ n/ 2 ⌋ X n 2 k n − 2 k ( n + 1) 3 (1) 2 k k k =0 2 k − k n Let ω = exp(2 πi/ ( n + 1)). The summand is ( ω + ω + 3) , which by the multinomial P P n n c k ( a − b ) expansion equals 3 ω . Since 0 ≤ | a − b | < n + 1, = ( n + 1) 1 . a = b a + b + c = n k =0 a,b,c Therefore the sum becomes n X X n n n − a − b n − 2 a ( n + 1) 3 1 = ( n + 1) 3 (2) a = b a, b, c a, a, n − 2 a a + b + c = n a =0 ⌊ n/ 2 ⌋ X n ! n − 2 a = ( n + 1) 3 (3) a ! a !( n − 2 a )! a =0 ⌊ n/ 2 ⌋ X n 2 a n − 2 a = ( n + 1) 3 , (4) 2 a a a =0 as claimed. The desired remainder is therefore 10 0 10 2 10 4 10 6 10 8 10 10 10 8 6 4 2 0 11 · 3 + 3 + 3 + 3 + 3 + 3 0 0 2 1 4 2 6 3 8 4 10 5 (5) 10 8 6 4 2 ≡ 11 · 3 + 3 · 90 + 3 · 10 · 6 + 3 · 10 · 20 + 3 · 45 · 70 + 52 (6) ≡ 91 (mod 100) (7) P 2 n − 1 j