Find the last three digits of `17^(256)dot`

Explanation

We have
`17^(256) = (17^(2))^(128) = (289)^(128) = (290-1)^(128)`
`:. 17^(256) = .^(128)C_(0)(290)^(128)-.^(128)C_(1)(290)^(127)+.^(128)C_(2)(290)^(126)-"....."`
`- .^(128)C_(125)(290)^(3)+.^(128)C_(126)(290)^(2)-.^(128)C_(127)(290)+1`
`=[.^(128)C_(0)(290)^(128) - .^(128)C_(1)(290)^(127)+.^(128)C_(2)(290)^(126)-"....."`
`-.^(128)C_(125)(290)^(3)]+.^(128)C_(126)(290)^(2) -.^(128)C_(127)(290)+1`
`=1000m+((128)(127))/(2)(290)^(2)-128xx290+1`
`= 1000 m + (128)(127)(290)(145)-128xx290-1`
`= 1000m+(128)(290)(127xx145-1)+1`
`=1000m+(128)(290)(18414)+1`
`=1000m+683527680+1`
`= 1000m+683527000+680+1`
`= 1000(m+683527)+681`
Hence, the last three digits of `17^(256)` are `6,8,1`. As a result, the last two digits of `17^(256)` are `8, 1` and the last digit of `17^(256)` is 1.