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Thread: Proof

  1. #1
    Junior Member
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    Proof

    91 five-digit numbers are written on a blackboard. Prove that one can find 3 numbers on the blackboard such that the sums of their digits are equal.

    I was thinking of writing the three numbers as $\displaystyle 10000a + 1000b + 100c + 10d + e $, $\displaystyle 10000f + 1000g + 100h + 10i + j $ and $\displaystyle 10000k + 1000l + 100m + 10n + o $, but how do I proceed? Also, I know that there are $\displaystyle 9*10^{4} $ different 5-digit numbers possible, but is that information going to help? This doesn't look like a permutations/combinations question to me...

    Please help immediately. It's for a friend... and my mathematical conscience is pricking me bad...

    Thanks!

    ILoveMaths07
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  2. #2
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    Hello, ILoveMaths07!

    It's easier than you think . . . the "pigeonhole principle".


    91 five-digit numbers are written on a blackboard.
    Prove that one can find 3 numbers on the blackboard
    such that the sums of their digits are equal.

    Five-digit numbers have the range: .$\displaystyle [10000, 99999].$

    Hence, their digit-sums have the range: .$\displaystyle [1,45]$

    . . That is, there are only 45 possible digit-sums.


    With 90 numbers of the board, it is possible they have two each
    . . of the 45 different digit-sums:$\displaystyle \{1,1,2,2,3,3,\,\hdots\, 45,45\}$


    The 91st number must duplicate one of the digit-sums.

    There will be 3 numbers with the same digit-sum.

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