Discrete Maths, Pigeonhole

drstress

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Attempt: I probably need to use pigeonhole principle… I tried to count the pairs of remainders their sum of difference is divisable by 20.
So, I found that, the remainders are (0,0) (1,19)...(9,11)(10,10) for the sum to be divisable by 20 and (0,0)(1,1)....(20,20) for their difference to be divisable by 20? There are total of 29 pairs.
 
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