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Sum divisable by 5.

  1. Nov 7, 2004 #1


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    The question asks me to prove that if I have 5 arbitrary natural numbers to show that one of the sums (in order) is divisible by 5. So say the numbers are [itex]a_1,a_2,a_3,a_4,a_5[/itex] some examples would be:

    [tex]a_1 + a_2 + a_3[/tex]


    [tex]a_3 + a_4[/tex]


    My first thought was to consider the remainder upon division 5 and then start identifying all the possible combinations where the sum is divisible by 5 and show there are no more left. However when starting this I realised this was actually a lot of work and there must be some simpler way. Can anyone else point me in a different direction?
  2. jcsd
  3. Nov 17, 2004 #2


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    I don't know what in order means so I hope it does not matter.
    consider the sums
    consider their remainders upon division by 5
    There are 5 possible remainders {0,1,2,3,4} and 5 sums so either a sum is divisible by 5 (remainder=0) or of two of the sums remainders are equal.
    if 2 sums remainders are equal the difference of the sums is itself a sum and is divisible by 5
    so for example if (a1,a2,a3,a4,a5,n,m,r are natural numbers)
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