Find (i) the last digit, (ii) the last two digits, and (iii) the last three digits of `17^(256)dot`

Explanation

We observed that `17^(2)=289=290-1`
Now `17^(256)=(17^(2))^(128)=(290-1)^(128)`
`=.^(128)C_(0)(290)^(128)-.^(128)C_(1)(290)^(127)+.^(128)C_(2)(290)^(126)+.^(128)C_(3)(290)^(125)` . . . . .`+.^(128)C_(126)(290)^(2)-128C_(127)(290)+1`
`=1000n+(128xx127)/(2)xx(290)^(7)-180xx290+1`, n being a positive integer
`=1000n+128xx290{(127)(145)-1}+1`
`=1000n+128xx290xx18414+1`
`=1000n+683527680+1`
`=1000(m+683527)+681`
Hence, the last three digits of `17^(256)` are 681 in that order