PUMaC 2022 · 代数（A 组） · 第 7 题
PUMaC 2022 — Algebra (Division A) — Problem 7
题目详情
- For a positive integer n ≥ 1, let a = ⌊ n + ⌋ . Given a positive integer N ≥ 1, let F n N 2 P 1 denote the set of positive integers n ≥ 1 such that a ≤ N . Let S = . As N goes to n N 2 a n n ∈F N 2 aπ infinity, the quantity S − 3 N tends to for relatively prime positive integers a, b . Given N b ∞ P 2 1 π that = , find a + b . 2 k 6 k =1 ∞
解析
- For a positive integer n ≥ 1, let a = ⌊ n + ⌋ . Given a positive integer N ≥ 1, let F n N 2 P 1 denote the set of positive integers n ≥ 1 such that a ≤ N . Let S = . As N goes to n N 2 a n n ∈F N 2 aπ infinity, the quantity S − 3 N tends to for relatively prime positive integers a, b . Given N b ∞ P 2 1 π that = , find a + b . 2 k 6 k =1 Proposed by Sunay Joshi Answer: 97 P 2 ∞ 1 1 1 π We claim that the desired limit equals , or equivalently , which yields an answer 2 k =1 16 k 96 of 97. √ 1 1 1 3 3 3 Note that a = k iff k ≤ n + < k + 1, or equivalently ( k − ) ≤ n < ( k + ) . Expanding, n 2 2 2 we find 3 3 1 3 3 1 3 2 3 2 k − k + k − ≤ n < k + k + k + 2 4 8 2 4 8 1 1 2 The upper and lower bounds differ by 3 k + . Note that if m is a real number with { m } > , 4 4 1 1 2 2 then there are exactly 3 k integers n in the interval [ m − (3 k + ) , m ). However if { m } ∈ (0 , ], 4 4 3 3 1 2 3 2 there are exactly 3 k + 1 integers in the interval. The upper bound of k + k + k + has 2 4 8 1 2 fractional part that is a multiple of . Thus there are 3 k + 1 values of n iff the fractional part 8 1 3 2 3 is exactly , namely when k + k is an integer and when k is divisible by 4. It follows that 8 2 4 2 2 there are 3 k values of n such that a = k if 4 does not divide k and 3 k + 1 values otherwise. n We may therefore rewrite the sum E as N ⌊ N/ 4 ⌋ N 2 X X X 3 k + 1 1 1 1 [4 | k ] E = − 3 N = = N 2 2 2 k k 16 k k =1 k =1 4 | k,k ≤ N 2 Sending N → ∞ , we find a limit of π / 96, as desired. ∞