Question

Find the last two digits of ${17}^{256}.$

Answer 1

${17}^{256}$

For finding last two digits we have to find the remainder of

${17}^{256}$

So, we apply binomial expansion ${17}^{256},$

$=(10+7{)}^{256}{=}^{256}{C}_0{10}^0{7}^{256-0}{+}^{256}{C}_1{10}^1{7}^{256-1}$

Higher terms which contains ${10}^2$ which are divisible by $100.$

So, ${}^{256}{C}_0{10}^0{7}^{256-0}{+}^{256}{C}_1{10}^1{7}^{256-1}$

We get last digit by last digit of ${}^{256}{C}_0{10}^0{7}^{256-0}$ which is last digit of $7^{256}$

$=1$

Second last digit is the second last digit of

${}^{256}{C}_1{10}^1{7}^{256-1}$

$=256\times 7^{255}\times 10$

$=6\times 3$

Second last digit $=8$ only the last digit (Because last digit of $7^{255}=3)$

So, the second last and last are $81.$