Find the remainder when $10^{20^{30}}$ is divided by $23$

Question

Find the remainder when $10^{20^{30}}$ is divided by $23$

I guess this question is fairly simple, but I just want to make sure I'm on the right track. My answer is shown below. If it is incorrect, please don't downvote it; I'll delete it soon after.

Edit: here's my work

Notice that $10^{2}\equiv 8\mod 23$, $8^2\equiv -5 \mod 23$, and $(-5)^2\equiv 2\mod 23$. Thus, $10^{16}\equiv 2\mod 23$ and $10^4\equiv -5\mod 23$ and so $10^{20}\equiv -10\mod 23\Rightarrow 10^{20^{30}}\equiv (-10)^{20^{29}}\equiv ...\equiv (-10)^{20}\equiv-10\mod 23\equiv 13\mod 23$.

mb I could use Fermat's Little Theorem. I guess I could make this more formal too.

How can we tell you if you are on the right track if you don't tell us what you have done? — fleablood, Oct 19 '19 at 04:38
Self-answering is not ok, when the question does not add anything new to the site. This is essentially a duplicate of this mother thread, already linked to by 170 other threads. — Jyrki Lahtonen, Oct 19 '19 at 04:55

score 3 · Accepted Answer · edited Oct 20 '19 at 00:51

3

By Fermat, $10^{22}\equiv 1\bmod 23,$ so it suffices to find $20^{30}\equiv (-2)^{30}\bmod 22.$ But $2^{6}=64\equiv-2\bmod 22\implies 2^{30}\equiv(-2)^5 \equiv 12\bmod 22.$ Therefore, the answer now is $\equiv10^{12}\bmod 23,$ which should be easily computable.

edited Oct 20 '19 at 00:51

answered Oct 19 '19 at 04:46

dezdichado

13,888

Find the remainder when $10^{20^{30}}$ is divided by $23$

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