Calculate $6^{1866}$ in $\mathbb{Z}_{23}$

Question

$Solution:$ Note that $1866=22\cdot 84 + 18$ then by Fermat's theorem

$$[6^{1866}]=[6^{22}]^{84}[6^{18}]=[1]^{84}[6^{18}]=[6^{18}]$$ Then $6^6=46656=2028\cdot 23 + 12$ It is true that $$[6^{1866}]=[6^6]^3=[-11]^3=[121][-11]=[72]=[3]$$

So $ 6 ^ {1866} $ is $3 $ in $ \mathbb {Z} _ {23} $, is that correct? Thank you for reading.

I didn‘t check the computations (which you can check with a calculator) but the logic is fine. — Qi Zhu, May 25 '20 at 05:48
The method is correct (e.g. see the dupe), and - of course - the arithmetic can be verified with a calculator. — Bill Dubuque, May 25 '20 at 05:57

score 1 · Accepted Answer · answered May 25 '20 at 05:54

1

Your work is correct.

You could have also said $6^{1866}\equiv6^{-4}\bmod23$.

$6^{-1}\equiv4$ and $4^4=256\equiv3\bmod 23.$

answered May 25 '20 at 05:54

J. W. Tanner

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Calculate $6^{1866}$ in $\mathbb{Z}_{23}$

1 Answers1