Question

Find the last three digits of ${17}^{256}$

Answer 1

The last $3$ digits of any number is remainder when the number is divided by $1000.$

So we just calculate the powers of $17$ up to $256$ as efficiently as possible and reduce them to modulo $1000$ as we go along.

${\left(17\right)}^2=289\left(mod1000\right)$

${17}^4={17}^{2^2}=83521\left(mod1000\right)$

${17}^8={17}^{4^2}={521}^2=271441=441\left(mod1000\right)$

${17}^{16}={17}^{8^2}={441}^2=481\left(mod1000\right)$

${17}^{32}={17}^{{16}^2}={481}^2=361\left(mod1000\right)$

${17}^{64}={17}^{{32}^2}={361}^2=321\left(mod1000\right)$

${17}^{128}={17}^{{64}^2}={321}^2=41\left(mod1000\right)$

${17}^{256}={17}^{{128}^2}={41}^2=681\left(mod1000\right)$

So the last digit is $681$