Is there an easier way to do this than finding cycles of different mods? Or can I just first do 333 (mod 17), it gives me 10. Then I could change all the 333's into 10s so it would be 10^10^10 and then do (10^10) first to get a remainder of 2 then 2^10 to get 4 in (mod 17), even though it might be time consuming?
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Is it $333^{(333^{333})}$ or $(333^{333})^{333}$ ? – TheSilverDoe Oct 22 '20 at 19:34
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Does this answer your question? Multiple Exponentiation in Modular Arithmetic – John Omielan Oct 22 '20 at 19:35
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You can't change all of the $333's$ into $10's$, that makes no sense. You can change only the bottom one. For example, $2^2$ and $2^5$ are not congruent mod $3$, even though $2\equiv 5$ (mod $3$). You can't change the power. – Mark Oct 22 '20 at 19:36
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No, you can't.$,$ – Oct 22 '20 at 19:37
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@JohnOmielan This is not really a duplicate. Bill Dubuque, aka the King of Duplicates, will be angry. – TheSilverDoe Oct 22 '20 at 19:50
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1Does this answer your question? How do I compute $a^b,\bmod c$ by hand? – Parcly Taxel Oct 22 '20 at 19:59
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Although $333\equiv10\bmod17$, it is not true that $333^{{333}^{333}}\equiv10^{{10}^{10}}\bmod17$.
Generally, $a\equiv b\bmod n$ does not imply that $c^a\equiv c^b\bmod n$.
Rather, $a\equiv b\bmod \phi(n)$ and $\gcd(c,n)=1$ implies $c^a\equiv c^b\bmod n$,
where $\phi(n)$ is Euler's totient function: $\phi(17)=16$ and $\phi(16)=8$.
Now $333\equiv13\bmod16$ and $333\equiv5\bmod8$, so $333^{333}\equiv13^5\bmod16$.
To reduce further, it helps to note that $13^4\equiv(-3)^4=81 \equiv1\bmod16$,
so $13^5\equiv13\bmod16$. Therefore, $333^{{333}^{333}}\equiv10^{{13}}\bmod17$.
Can you take it from here?
J. W. Tanner
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