How can one find out the value of $43^{1234567890} \mod 22\;?$
Can I just say that because $123456890$ is an even number I can calculate $43^2 \mod 22$, which is $1$?
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4Well, I'd point out first that $43 \equiv -1 \mod 22$. And $-1$ raised to an even power equals 1 – Ronald Oct 29 '18 at 13:08
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https://math.stackexchange.com/questions/289137/calculate-201234567-mod-251?rq=1 – Darío A. Gutiérrez Oct 29 '18 at 13:10
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Yes that’s fine, it suffices use that
$$a \equiv b \mod m \implies a^n \equiv b^n \mod m$$
and in that case $43 \equiv -1 \mod {22}$.
user
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It turns out to be true, but the way you phrase it seems a bit wrong.
So, just to avoid any confusion, you can calculate $43^2 \pmod {22}$ but only because $43 \equiv -1 \pmod {22}$ and ${(-1)}^n = {(-1)}^2$ for every even n.
In other words, $42^{1234567890} \equiv 42^2 \pmod {22}$ would have been false
F.Carette
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1$43^{1234567890} \equiv 43^2 \pmod{22}$ is perfectly correct. Whether an instructor would accept it without further justification is another story. – fkraiem Oct 29 '18 at 13:25
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@fkraiem It is indeed totally correct, I just wanted to be sure OP only use it when they're allowed. As for the instructor part, it's mainly opinion biaised, but my former math teachers wouldn't have accepted it. – F.Carette Oct 29 '18 at 13:30