Sum divisable by 5.

1. Nov 7, 2004

Zurtex

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 $a_1,a_2,a_3,a_4,a_5$ some examples would be:

$$a_1 + a_2 + a_3$$

$$a_4$$

$$a_3 + a_4$$

etc..

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. Nov 17, 2004

lurflurf

I don't know what in order means so I hope it does not matter.
consider the sums
a1
a1+a2
a1+a2+a3
a1+a2+a3+a4
a1+a2+a3+a4+a5
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)
a1+a2=5m+r
a1+a2+a3+a4=5n+r
5|a3+a4