The last two digits of the number `3^(400)` are:
(A) `81` (B) `43` (C) `29` (D) `01`

Explanation

We have , `3^(400) = (3^(2))^(200) = (9)^(200) = (10 -1)^(200)`
` = (10)^(200) - ""^(200)C_(1) (10)^(199) + ""^(200)C_(2) (10)^(198) - ""^(200)C_(3) (10)^(197)+ ..+ ""^(200)C_(199) (10)^(2) - ""^(200)C_(199)(10) + 1`
` = 100 mu = ""^(200)C_(199) (10) + 1 " where" mu in I `
` 100 mu - ""^(200)C_(1) (10) + 1 = 100 mu - 2000 + 1`
` = 100 (mu - 20 ) + 1 = 100 p + 1 ` , where p is an integer
Hence , the last two digits of ` 3^(400) " is " 00 + 1 = 01 ` .