Last Digit of a Power

In order to get the last two digits of $3^{243}$ in base 8, we first take the number $\mod{64}$ , since $8^{2} = 64$ , in base 10. Also note that $3^{2} = 9 = 8+1$ , so we can easily reduce this exponent by using the binomial theorem.

$\begin{aligned} 3^{243} \mod{64} &\equiv 3\left(9^{121}\right) \mod{64} \\ &\equiv 3\left(8+1\right)^{121} \mod{64} \\ &\equiv 3\left(\binom{121}{1}(8)^{1}(1)^{120} + \binom{121}{0}(8)^{0}(1)^{121} \right) \mod{64} \\ &\equiv 3\left(121(8) + 1\right) \mod{64} \\ &\equiv 3(9) \mod{64} \\ &\equiv 27 \mod{64} \end{aligned}$

We have now found that we have found the base 10 representation of the last two digits of the power in base 8, so we convert the number to base 8, which is $\boxed{33}$ .

$\text{215475607473713760163677411770041767434032073214605243331671250157070666002444331545125411500242636614741673350026147474265634133}_8$

First off, I know this question is a test to tell who is not enjoying the holidays ;D

The solution is straightforward. Calculate the number, modulo $8^2=64$ . You will probably need Euler's theorem for that. Then the result $27$ , which is a base $10$ number, can be converted to base $8$ to get $33$ .