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HMMT 二月 2017 · COMB 赛 · 第 3 题

HMMT February 2017 — COMB Round — Problem 3

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. There are 2017 jars in a row on a table, initially empty. Each day, a nice man picks ten consecutivejars and deposits one coin in each of the ten jars. Later, Kelvin the Frog comes back to see that N ofthe jars all contain the same positive integer number of coins (i.e. there is an integer d > 0 such that
    N of the jars have exactly d coins). What is the maximum possible value of N ?
解析

英文解析

  1. There are 2017 jars in a row on a table, initially empty. Each day, a nice man picks ten consecutivejars and deposits one coin in each of the ten jars. Later, Kelvin the Frog comes back to see that N ofthe jars all contain the same positive integer number of coins (i.e. there is an integer d > 0 such that
    N of the jars have exactly d coins). What is the maximum possible value of N ?
    Proposed by: Kevin Sun
    Answer: 2014
    Label the jars 1 , 2 , . . . , 2017 . I claim that the answer is 2014 . To show this, we need both a constructionand an upper bound. For the construction, for 1 ≤ i ≤ 201, put a coin in the jars 10 i + 1 , 10 i +
    2 , . . . , 10 i + 10 . After this, each of the jars 1 , 2 , . . . , 2010 has exactly one coin. Now, put a coin in eachof the jars 2008 , 2009 , . . . , 2017 . Now, the jars 1 , 2 , . . . , 2007 , 2011 , 2012 , . . . , 2017 all have exactly onecoin. This gives a construction for N = 2014 (where d = 1).
    Now, we show that this is optimal. Let c , c , . . . , c denote the number of coins in each of the jars.
    1 2 2017
    For 1 ≤ j ≤ 10, defines = c + c + c + . . . .
    j j j +10 j +20
    Note that throughout the process, s = s = · · · = s . It is also easy to check that the sums s , s , . . . , s
    1 2 j 1 2 7
    each involve 202 jars, while the sums s , s , s each involve 201 jars.
    8 9 10
    Call a jar good if it has exactly d coins. If there are at least 2015 good jars, then one can check that it isforced that at least one of s , s , . . . , s only involves good jars, and similarly, at least one of s , s , s
    1 2 7 8 9 10
    only involves good jars. But this would mean that 202 d = 201 d as all s are equal, contradiction. i