Calculate last two digit; last three digit of `(17)^256`

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 `
Foe last three digits ,
`17^(256) = (290)^(3) [ ""^(128)C_(0) (290)^(125) - ""^(128)C_(1) (290)^(124) + ""^128C_(2) (290)^(123) - ... - ""^(128)C_(125) (1)] + ""^(128)C_(126) (290)^(2) -""^(128)C_(127) (290) + 1`
` = 1000m + ""^(128)C_(126) (290)^(2) - ""^(128)C_(127) (290) + 1`
where , m is an integer
`= 1000 m + ""^(128)C_(2) (290)^(2) - ""^(125)C_(1) (290)+ 1`
` 1000m + ((128)(127))/(2) (290)^(2) - 128 xx 290 + 1`
` = 1000 m + (128) (127)(290)(145)- (128) (290) + 1`
` = 1000 m + (128) (290) (127xx 145- 1 ) + 1`
` = 1000m + (128) (290) (18414)+ 1`
` 1000m + 683527000 + 680 + 1`
` = 1000 (m + 683527) + 681 `
` therefore ` Last three digits = ` 000 + 681 = 681 `