Question

Find the last digit of ${17}^{256}.$

Answer 1

${17}^{256}$

Note:- Multiplicity of $7$ is $4$ (Means after every $4th$ power of $7$ the last digit repeat that is

$7^1=7,$

$7^4=2401$

$7^5=16807$

$7^6=117649$

Last digit is $9.$

So, we have powers last digits $\{7,9,3,1\}$

So, in ${17}^{256}$ last digit is same as $7^{256}=7^{4\times 63+4}=7^4$

So, the last digit is $1.$