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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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