<s>[INST] How to find the last two digit of $7^{81}$? Could any one tell me how to find the last two digit of $7^{81}$? I have succeeded in finding the last digit only which is $7$.
Any group theoretic approach or any other approach is welcome.
 [/INST] An alternative method of determining the value that does not rely on calculating $\phi(100)$ is to perform the calculation in modular arithmetic using efficient exponentiation.
$81=64+16+1$, so here's the process (all lines are mod 100):

$$7 \equiv 7\\7^2 \equiv 49\\7^4 \equiv 49^2 \equiv 1\\7^5 \equiv 1\cdot 7 \equiv 7\\7^{10} \equiv 7^2 \equiv 49\\7^{20} \equiv 49^2 \equiv 1\\7^{40} \equiv 1^2 \equiv 1\\7^{80} \equiv 1^2 \equiv 1\\7^{81} \equiv 1\cdot 7 \equiv 7$$

This approach works for essentially any set of base, exponent, and modulo number.
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