If integers a and b have remainders r and s on division by n, show that a+b and r+s have have the same remainder on dividing by n.

(a-r)/n=integer (b-s)/n=another integer adding the two {a+b-(r+s)}/n=some other integer

just wait i will modify it a bit

means when a+b is divided by n we get r+s as remainder if r+s is less than n, That means r+s when r+s <n,, r+s divided by n , r+s is the remainder. But if it is greater than n then on dividing a+b by n, r+s/n 's remainder will be its remainder . Sorry am not able to explain it better, hope u understood...

I can see where u are going although it might not be that clear for others..

Let a = q1n+r and b = q2n +s and let t be the remainder from (r+s)/n ie r+s = q3n +t Then add the first 2 equations....

yep looks like u got it

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