Question

Prove the following: Show that for any given 92 integers there exist two of them whose sum, or else whose difference, is divisible by 180.

Answer 1

We know that e given any n+2 integers, there exist tero of them whose sum on else difference is divisible by 2n. 2n consider the congruence classes Rod modulo an which we will denote by 202, put tero congruence classes P and q in the same Piseinhole if p=2n-a 2n-1} mod an. n are There are (nal) such Pigeion hole because o O and ooo on their own and the remaining 2n-2. congruena congruence classes are paired in (ned) pairs (1,2n-I ) (2, 2n-2), ---- enas, nad)

There se are cnt2) integers s and nid Pigeon holes there must be at least in the same Pigeonhole tero integers a,b ES neither : then a= b. mod an » aab-o mod an or, a a 2n-b med 2n = a +b =0 mod 2n. Here we consider nago and 2.90-180 Hence, 90+2)=92 Thus for any given 92 integers there exist to of them whose else whose difference is divisible by 180. sum or