Find the last 3 digits of 101^99.
101^99 = (100 + 1)^99
Binomial expansion
C(99,0)*100^99 + ...........+ C (99.98) (100) + C(99,99) *1
The last two terms will give us the last 3 digits
C(99,98) * 100 + C(99,99) * 1 =
9900 + 1 =
901
