How can I prove in generality that for every power of 9, 9^k mod 10 will be either 9 or 1? I know I can give a few cases like 9^2 = 81 mod 10 = 1 etc, but is there a more general way to prove?
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1There absolutely is. First, think about the usual integers: can you show that every power of $-1$ will be either $-1$ or $1$? Now, do you see a connection between "$-1$ in $\mathbb{Z}$" and "$9$ in mod-$10$-world"? (This is a bit more abstract than is necessary, but I think helps provide a satisfying "story" in addition to a proof.) – Noah Schweber Nov 06 '21 at 05:10
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Induction with two basis cases? – Adam Rubinson Nov 06 '21 at 05:10
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$9^n \equiv (-1)^n$ (mod $10$). So if n is even, it's equivalent to $1$ and if n is odd, it's equivalent to $-1$(mod $10$) that is 9.
Mina
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Just using induction will give you an answer.
If $n=1$, $9^n\equiv1\text{ or }9\pmod{10}$ is so trivial.
Assuming $n=i$, then $9^n\equiv1\text{ or }9\pmod{10}$. Then $9^{i+1}=9\times9^i\equiv9\text{ or } 1\pmod{10}$.
So, by induction, for all $n\in\mathbb N$, $9^n\equiv1\text{ or }9\pmod{10}$.
MH.Lee
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