The sum of last 3 digits of `3^100` is

Explanation

As we know that `3^100 = 9^50 = (10-1)^50`

by using binomial expansion we can write

`(10-1)^50 = .^50C_0 10^50- .^50C_1 10^49 + .. .. .^50C_48(10)^2 - .^50C_49(10)1 + .^50C_50`

`= (50 xx 49)/2 xx 100 - 50 xx 10 + 1`

`= 122500 - 500 +1`

`= 122001`

So, sum of last 3 digit is `= 0 + 0 =1 = 1`

option A is correct.