Find the last two digits of $ 7^{81} ?$ I came across the following problem and do not know how to tackle it.  

Find the last two digits of $ 7^{81} ?$  

Can someone point me in the right direction? Thanks in advance for your time.

We use the technique found in our answer to
$\quad$ last two digits of $14^{5532}$?
but now express it with more succinctness.

The integer $81$ has a binary representation of $1010001$.
We have the operator
$$ n \text{ mod 100} \mapsto n^2 \text{ mod 100}$$
and so we need to successively apply it at most $\text{len}(1010001) - 1$ (i.e. $6$ times) to the residue class $[7]$,
$$ [7] \mapsto [49] \mapsto [1]$$
Since we hit $[1]$ our work is done; the answer is
$$ (1 \cdot [1]) (\text{from } 2^6) \times  (1 \cdot [1]) (\text{from } 2^4) \times  (1 \cdot [7]) (\text{from } 2^0) = [7]$$
