1.

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

Answer» 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.


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