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

Options

A. `1`
B. `2`
C. `3`
D. `4`

Answer

A

Explanation

`(a)` We have
`3^(100)=(3^(4))^(25)`
`=(81)^(25)`
`=(80+1)^(25)`
`=^(25)C_(0)*(80)^(25)+^(25)C_(1)*(80)^(24)+....+^(25)C_(22)(80)^(3)+^(25)C_(23)(80)^(2)+^(25)C_(24)(80)+^(25)C_(25)`
`=10^(3)['^(25)C_(0)8^(25)xx10^(22)+^(25)C_(1)xx8^(24)xx10^(21)+...+^(25)C_(2)xx8^(3)]+(25xx24)/(2)xx(80)^(2)+25xx80+1`
`=10^(3)m+1920000+2000+1`, where `m in N`
Thus, last three digits of `3^(100)` are `001`,