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HMMT 二月 2004 · GEN2 赛 · 第 6 题

HMMT February 2004 — GEN2 Round — Problem 6

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. a and b are positive integers. When written in binary, a has 2004 1’s, and b has 2005 1’s
    (not necessarily consecutive). What is the smallest number of 1’s a + b could possiblyhave?
解析

英文解析

  1. a and b are positive integers. When written in binary, a has 2004 1’s, and b has 2005 1’s
    (not necessarily consecutive). What is the smallest number of 1’s a + b could possiblyhave?
    Solution: 1
    Consider the following addition:
    111 · · · 100 · · · 01 + 11 · · · 11 = 1000 · · · · · · · · · 00
    By making the blocks of 1’s and 0’s appropriately long, we can ensure that the addends 2
    4008 2005
    respectively contain 2004 and 2005 1’s. (To be precise, we get a = 2 − 2 + 1 and
    2005
    b = 2 − 1.) Then the sum has only one 1. And clearly it is not possible to get anyless than one 1.