HMMT 二月 2019 · ALGNT 赛 · 第 9 题
HMMT February 2019 — ALGNT Round — Problem 9
题目详情
- Tessa the hyper-ant has a 2019-dimensional hypercube. For a real number k , she calls a placement of 2019 nonzero real numbers on the 2 vertices of the hypercube k -harmonic if for any vertex, the sum of all 2019 numbers that are edge-adjacent to this vertex is equal to k times the number on this vertex. ∑ Let S be the set of all possible values of k such that there exists a k -harmonic placement. Find | k | . k ∈ S ∞
解析
- Tessa the hyper-ant has a 2019-dimensional hypercube. For a real number k , she calls a placement of 2019 nonzero real numbers on the 2 vertices of the hypercube k -harmonic if for any vertex, the sum of all 2019 numbers that are edge-adjacent to this vertex is equal to k times the number on this vertex. ∑ Let S be the set of all possible values of k such that there exists a k -harmonic placement. Find | k | . k ∈ S Proposed by: Yuan Yao Answer: 2040200 By adding up all the equations on each vertex, we get 2019 S = kS where S is the sum of all entries, so k = 2019 unless S = 0. In the latter case, by adding up all the equations on a half of the cube, we get 2018 S − S = kS where S is the sum of all entries on that half of the cube, so k = 2017 unless S = 0. In the latter case (the sum of all entries of any half is zero), by adding up all the equations on a half of the half-cube, we get 2017 S − 2 S = kS , so k = 2015 unless S = 0. We continue this chain of casework until we get that the sum of every two vertices connected by unit segments is zero, in which case we have k = − 2019. This means that k can take any odd value between − 2019 and 2019 inclusive, so the 2 sum of absolute values is 2 · 1010 = 2040200. 2019 To achieve these values, suppose that the vertices of the hypercube are { 0 , 1 } and that the label of x x x 1 2 2019 ( x , x , . . . , x ) is a a . . . a for constants a , a , . . . , a ∈ {− 1 , 1 } , then it is not difficult to 1 2 2019 1 2 2019 1 2 2019 see that this labeling is ( a + a + · · · + a )-harmonic for any choice of a ’s, so this can achieve all 1 2 2019 i odd values between − 2019 and 2019 inclusive. ∞