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Fermat's little theorem states that if $p$ is a prime number and $\operatorname{gcd}(a,p)=1$, then $a^{p-1} -1$ s a multiple of $p$.

Here's my attempt to find the remainder when $5^{99999}$ is divided by 53.

$5^{99999} = 5^{52}\equiv 1\pmod{53} \implies 5^{99999}\equiv5^{99999\bmod52}=5^{3}.$

Is this the correct approach? How do I proceed from here?

Emilycodes
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    Your approach is fine. Now can you calculate $5^3 \pmod {53}$? – player3236 Feb 22 '21 at 17:54
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    Your approach is basically OK, but the opening, $5^{99999}=5^{52}$, is neither correct nor necessary. (The only other thing I'd add is "proof" that $99999\equiv3$ mod $52$, e.g., a line along the lines of "since $99999=1923\cdot52+3$.") As for the next step, note that $5^3=125$ is bigger than $53$, so you still need to find a remainder that's less than $53$. – Barry Cipra Feb 22 '21 at 17:59
  • Thank you for the help. – Emilycodes Feb 22 '21 at 18:12

1 Answers1

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You have just proved that $5^{99999} \equiv 5^3$ $(mod\space53)$. They have the same remainder when divided by $53$. So now you just need to find the remainder of $125$ when divided by $53$.

There is also an extra $5^{99999}$ at the beginning of the third line

Mariano Juncal
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