PUMaC 2011 · 数论（B 组） · 第 7 题

PUMaC 2011 — Number Theory (Division B) — Problem 7

[ 7 ] Find the sum of all positive integers k with k ≤ 1000 such that there exists an integer n > k that satisfies ( ) 2 n k ∑ ∑ 3 i = 3 i . i = k +1 i =1

解析

Rewrite i as i − i = − . Also, rewrite the right side as 3 . 4 4 4 i =1 i =1 i = k +1 2 2 2 2 Therefore, n ( n +1) = 4 k ( k +1) . Taking the square root of both sides shows that n ( n +1) = 2 k ( k + 1) (the negative case is not possible for positive n and k ). Multiply both sides by 2 2 4 and add 1 to show that (4 n + 4 n + 1) = 2(4 k + 4 k + 1) − 1, which is equivalent to 2 2 2 2 − 1 = (2 n + 1) − 2(2 k + 1) . Letting x = 2 k + 1 and y = 2 n + 1 shows that − 1 = y − 2 x , with ( x, y ) = (1 , 1) as the base solution. √ √ 2 j +1 All solutions can be obtained by computing (1 + 2) = y + x 2 for values of j such ( ) ( ) ( ) 2 j +1 2 j +1 2 j +1 j that x ≤ 2001. In this equation, x = + 2 + . . . + 2 , which can be seen by 1 3 2 j +1 √ √ expanding the expression for y + x 2 and equating coefficients of 2. Note that for j = 4, x = 9 + 168 + 504 + 288 + 16 = 985. For j ≥ 5, x ≥ 11 + (11(5)(3))(2) + (11(2)(3)(7))(4) + (11(10)(3))8 + (11(5))16 + 32 > 2001, so there are at most 4 solutions for x > 1 and x ≤ 2001. Note that j = 1 , 2, and 3 correspond to x = 5 , 29, and 7 + 70 + 84 + 8 = 169. These correspond to k = 2 , 14 , 84, and 492, so the desired sum is 592 . √