30^99 = ( 31 - 1 )*99
61^100 = ( 62 -1 )^100
The binomial expansion of the first will have all terms divisible by 31 in every term except the last which = (-1)^99 = -1
The binomial expansion of the second will have some power of 62 in every term (each divisible by 31) except the last term which is (1)^100 = 1
So....adding the expansions will give us all terms that will be divisible by 31 plus -1 + 1 = 0
So the remainder of this sum divided by 31 = 0