If the number is ` 17^(256)` , find the last digit

Explanation

Since , `17^(256) = (17^(2))^(128) = (289)^(128) = (290-1) ^(128)`
` therefore 17^(156) = ""^(128)C_(0)(290)^(128) - ""^(128)C_(1) (290)^(127) + ""^(128)C_(2) (290)^(126)`
`- ""^(128)C_(3) (290)^(125) +...- ""^(128)C_(125) (290)^(3) + ""^(128)C_(126) (290)^(2) - ""^(128)C_(127) (290) + 1 `
For last two digits ,
`17^(256) = (290)^(2) [ ""^(128)C_(0) (290)^(126) - ""^(128)C_(1) (290)^(125) + ""^128C_(2) (290)^(124) - ... + ""^(128)C_(128) (1)] - ""^(128)C_(127) (290) + 1`
` = 100 m - ""^(128)C_(127) (290) + 1` , where m is an integer .
` = 100 m - ""^(128)C_(1) (290) + 1 = 100 m - 128xx 290 + 1`
` = 100 (m - 384) + 1281`
`= 100 n + 1281` , where n is an integer
` therefore ` Last two digits = 00 + 81 = 81 .