HMMT 二月 2023 · ALGNT 赛 · 第 9 题
HMMT February 2023 — ALGNT Round — Problem 9
题目详情
- For any positive integers a and b with b > 1, let s ( a ) be the sum of the digits of a when it is written b in base b . Suppose n is a positive integer such that b log n c b log n c 23 20 (⌊ ⌋) (⌊ ⌋) ∑ ∑ n n s = 103 and s = 115 . 20 23 i i 23 20 i =1 i =1 Compute s ( n ) − s ( n ) . 20 23 2 πi/ 99 2 πi/ 101
解析
- For any positive integers a and b with b > 1, let s ( a ) be the sum of the digits of a when it is written b in base b . Suppose n is a positive integer such that b log n c b log n c 23 20 (⌊ ⌋) (⌊ ⌋) ∑ ∑ n n s = 103 and s = 115 . 20 23 i i 23 20 i =1 i =1 Compute s ( n ) − s ( n ) . 20 23 Proposed by: Luke Robitaille, Raymond Feng Answer: 81 Solution: First we will prove that ( ) ∞ ⌊ ⌋ ∑ n s ( n ) = n − ( a − 1) . a i a i =1 If n = ( n n · · · n n ) , then the digit n contributes n to the left side of the sum, while it contributes k k − 1 1 0 a i i i i − 1 i − 2 1 0 n ( a − ( a − 1)( a + a + · · · + a + a )) = n i i to the right side, so the two are equal as claimed. Now we have ∞ (⌊ ⌋) ∑ n 103 = s 20 i 23 i =1 ⌊ ⌋ ∞ ∞ ⌊ ⌋ i ∑ ∑ n b n/ 23 c = − 19 i j 23 20 i =1 j =1 ∞ ∞ ∞ ⌊ ⌋ ⌊ ⌋ ∑ ∑ ∑ n n = − 19 , i j i 23 20 · 23 i =1 i =1 j =1 ⌊ ⌋ ⌊ ⌋ b n/p c n where we have used the fact that = for positive integers n, p, q . Similarly, q pq ∞ ∞ ∞ ⌊ ⌋ ⌊ ⌋ ∑ ∑ ∑ n n 115 = − 22 . j j i 20 20 · 23 j =1 i =1 j =1 Let ∞ ∞ ∞ ∞ ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ ∑ ∑ ∑ ∑ n n n A = , B = , and X = . j i j i 20 23 20 · 23 j =1 i =1 i =1 j =1 Then we have 103 = B − 19 X and 115 = A − 22 X . Thus, we have ( ) ∞ ∞ ⌊ ⌋ ⌊ ⌋ ∑ ∑ n n s ( n ) − s ( n ) = n − 19 − n − 22 20 23 j i 20 23 j =1 i =1 = 22 B − 19 A = 22(103 + 19 X ) − 19(115 + 22 X ) = 22 · 103 − 19 · 115 = 81 . Remark. The value n = 22399976 satisfies both equations, so a valid solution to the system exists. It seems infeasible to compute this solution by hand. 2 πi/ 99 2 πi/ 101