HMMT 二月 2015 · 冲刺赛 · 第 11 题

HMMT February 2015 — Guts Round — Problem 11

[ 6 ] Find √     2000000 ∞ 1 + ∑  k  4   , 2 k =0 where ⌊ x ⌋ denotes the largest integer less than or equal to x .

解析

[ 6 ] Find √     2000000 ∞ 1 + ∑  k  4   , 2 k =0 where ⌊ x ⌋ denotes the largest integer less than or equal to x . Answer: 1414 The k th floor (for k ≥ 0) counts the number of positive integer solutions to k 2 6 k 2 4 (2 x − 1) ≤ 2 · 10 . So summing over all k , we want the number of integer solutions to 4 (2 x − 1) ≤ 6 2 · 10 with k ≥ 0 and x ≥ 1. But each positive integer can be uniquely represented as a power of 2 √ 3 times an odd (positive) integer, so there are simply ⌊ 10 2 ⌋ = 1414 solutions.