Question

Find the last two digits of ${19}^{266}$

Answer 1

${19}^{266}=({19}^2{)}^{133}$

Now ${19}^2$ ends in 61 (${19}^2=361$).

Therefore, we need to find the last two digits of ${61}^{133}$.
Once the number is ends in $1$, we can straight away get the last two digits with the help of the previous method.

The last two digits are $81(6\times 3=18$ so the tenth digit will be $8$ and last digit will be $1)$.