How do you do this mod question?

2 of a, b and c must have the same (mod 2), since it is only either 0mod2 or 1mod2. you'll have at least 2 of the same parity (odd / even), whichever it is, the product will be divisible by 2.

0mod3, 1mod3 or 2mod3.
the last one will match either, if the first 3 do not.
0mod3 = 0mod3 - 0mod3
= 1mod3 - 1mod3
= 2mod3 - 2mod3

pigeonhole?

generalization :
product of differences of n integers (there are nC2 = n(n-1)/2 of those) is divisible by n-1.

out of a b c we know either at least 2 numbers are even or odd.

so one of the diffrences is is divsible by 2 so (a-b)(b-c)(c-a)

taking more formally say a-b is not even

then one is odd and another even

with out loss of generality say a is odd and b even

now c is either even in which case b-c is divisible by 2 or c is odd then c-a is divisible by 2

in which case (a - b)(b - c)(c - a) is



for the second part

working is mod 3

in the worst case 3 numbers have different remainder and hence 4th one has to be same on one of the 3 and hence one of the differences is divsible by 3


it can be generalised

out of a set on n integers product of all the difference of each pair of numbers is divisible by n-1