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I tried to solve this question but without a success. for every prime number $$p\ge7 $$ and every $$n \in \mathbb N$$ : $$10^{n(p-1)}\equiv 1 (\text{mod }9p) $$

I tried to use Euler theorem. but It was't helpful. I would like to get help with this question thanks

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$10^{n(p-1)}\equiv 1^{n(p-1)}\equiv 1 \mod 9$ and $10^{n(p-1)}\equiv (10^{(p-1)})^n\equiv 1^n\equiv 1 \mod p$ (second congruence is Fermat little theorem, which we can use as $p\nmid 10$). So both $9$ and $p$ divide $10^{n(p-1)}-1$, so $9p$ divides $10^{n(p-1)}-1$, so $10^{n(p-1)}\equiv 1\mod 9p$.

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Note that by the $10^{n(p-1)} \equiv 1 \pmod{9}$, since $10\equiv 1 \pmod{9}$ and $10^{n(p-1)}=(10^{p-1})^n\equiv 1\pmod{p}$ by Fermat's little theorem which applies since $p\ge 7$, so $p\nmid 10$. Then by the Chinese Remainder Theorem, since $9$ and $p$ are relatively prime (since $p\ne 3$), we have $10^{n(p-1)}\equiv 1\pmod{9p}$.

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