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find the remainder of

$$3^{20}$$

divided by 7. So we know what number its congruent mod 7 too.

I know

$$3^3 \equiv (-1) \mod 7$$

Thus

$$3^{18} \equiv 1 \mod 7$$

multiplying through by $3^2$ we get

$$3^{20} \equiv 3^2 \mod 7 \equiv 2 \mod 7$$

So the answer is $2$?

Snaw
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homosapien
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    You're correct. One quite minor, slightly easier, thing is you could have used that $3^6 \equiv 1 \pmod{7}$, which comes from Fermat's little theorem. – John Omielan Mar 20 '22 at 22:16
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    This is a special case of the method described in the linked dupes. Please note that for solution-verification questions to be on-topic you should state precisely what step you doubt, and why you doubt it. – Bill Dubuque Mar 20 '22 at 22:59

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