PUMaC 2023 · 数论（A 组） · 第 2 题
PUMaC 2023 — Number Theory (Division A) — Problem 2
题目详情
- A number is called good if it can be written as the sum of the squares of three consecutive positive integers. A number is called excellent if it can be written as the sum of the squares 2 2 2 of four consecutive positive integers. (For instance, 14 = 1 + 2 + 3 is good and 30 = 2 2 2 2 1 + 2 + 3 + 4 is excellent.) A good number G is called splendid if there exists an excellent number E such that 3 G − E = 2025. If the sum of all splendid numbers is S , find the remainder when S is divided by 1000.
解析
- A number is called good if it can be written as the sum of the squares of three consecutive positive integers. A number is called excellent if it can be written as the sum of the squares 2 2 2 of four consecutive positive integers. (For instance, 14 = 1 + 2 + 3 is good and 30 = 2 2 2 2 1 + 2 + 3 + 4 is excellent.) A good number G is called splendid if there exists an excellent number E such that 3 G − E = 2025. If the sum of all splendid numbers is S , find the remainder when S is divided by 1000. Proposed by Sunay Joshi Answer: 447 2 2 2 2 Any good number can be written as ( n − 1) + n + ( n + 1) = 3 n + 2 for some n ≥ 2. 2 2 2 2 Similarly any excellent number can be written as ( m − 1) + m + ( m + 1) + ( m + 2) = 2 2 (2 m + 1) + 5 for some m ≥ 2. Therefore, a good number 3 n + 2 (with n ≥ 2) is excellent 2 2 iff there exists m ≥ 2 such that 3(3 n + 2) − ((2 m + 1) + 5) = 2025. Rearranging, this 2 2 is equivalent to 9 n − (2 m + 1) = 2024. By the difference of squares, this becomes (3 n − 3 (2 m + 1))(3 n + (2 m + 1)) = 2 · 11 · 23. Since the two factors add to the even number 6 n , it is clear that each must be even. The possible factor pairs ( x, y ) with x < y and x, y > 0 are thus (2 , 4 · 11 · 23), (4 , 2 · 11 · 23), (2 · 11 , 4 · 23), and (4 · 11 , 2 · 23). These correspond to ( n, m ) = (169 , 252) , (85 , 125) , (19 , 17) , (15 , 0). Since we require m, n ≥ 2, only the first three pairs are valid solutions. It follows that n ∈ { 169 , 85 , 19 } , so the set of splendid numbers is 2 2 2 2 2 2 { 3 · 169 + 2 , 3 · 85 + 2 , 3 · 19 + 2 } . Thus S = (3 · 169 + 2) + (3 · 85 + 2) + (3 · 19 + 2), which is congruent to 447 (mod 1000).