$$7^{101} \mod 5 = 49^{100} \mod 5 = (49^2)^{50} \mod 5$$
What is the next step to find solution?
Regards
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1$\large \bmod 5!:\ 7\equiv 2,\ \ \color{#c00}{2^4\equiv 1},\Rightarrow, 2^{4n}\equiv (\color{#c00}{2^4})^n\equiv \color{#c00}1^n\equiv 1\ \ $ – Bill Dubuque Dec 14 '18 at 16:18
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2How did you get from $7^{101}$ to $49^{100}$? – farruhota Dec 14 '18 at 16:27
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You are wrong in saying $7^{101}\equiv 49^{100}$ you need $7\cdot 49^{50}$ - just take care when manipulating the powers. – Mark Bennet Dec 14 '18 at 16:27
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1Not every tag which starts with the letters "mod" is about modular arithmetic. I'm surprised you didn't add (modular-forms) in addition to (modules) and (model-theory)! – Alex Kruckman Dec 14 '18 at 18:55
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This is also a duplicate of a mother thread where the techniques for answering questions of this type have been collected. Shame on experienced users who choose to indulge in on-site-plagiarism. – Jyrki Lahtonen Dec 15 '18 at 06:13
3 Answers
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There is something wrong in your first step, indeed
$$7^{101} \not \equiv 49^{100} \mod 5 \ldots $$
More simply use that
$$7^2 \equiv 49 \equiv -1 \pmod 5$$
therefore
$$7^{101} \equiv 7\cdot7^{100} \equiv 7\cdot (7^{2})^{50} \equiv \,? \pmod 5$$
user
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By Carmichael function if $(a,m)=1$, then $a^{k \lambda (M)}\equiv 1 (mod m)$. That $\lambda (5)=4$. So $7^{4k}\equiv 1 (mod 5)$.
$7^{100}\equiv 1 (mod 5)$ then $7^{101}\equiv 2 (mod 5)$.
yavar
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What if you do not see a cute way to do this?
There is a more systematic way to do it.
First, $7^x\pmod 5$ is periodic in $x$. Compute a few of them until you see what the pattern is. Then it should be easy to find $7^{101}\pmod 5$.
GEdgar
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