Question

Find the last two digits of ${64}^{236}$.

Answer 1

${64}^{236}=(2^6{)}^{236}=2^{1416}=(2^{10}{)}^{141}\times 2^6={24}^{141}$ (24 raised to odd power)
Now, those numbers which are not in the form of $2^n$ can be broken down into the form $2^n\times$ odd number.

We can find the last two digits of both the parts separately.