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3 hours ago history became hot network question
7 hours ago comment added Robert Shore To be precise, don't assume $13 \nmid x$. Prove it by contradiction. Assume toward a contradiction that there is a solution with $13 \mid x$. But then also $13 \mid x^{101}$, which means that $x^{101} \equiv 0 \neq 5 \pmod{13}$. That contradiction shows that our assumption that $13 \mid x$ has to be false.
9 hours ago answer added Shaun timeline score: 3
11 hours ago answer added Bill Dubuque timeline score: 3
11 hours ago comment added Malady I am not sure I understand your concern. You were asked to find a solution, and you found one. You were not asked to prove that this solution was unique. If you were, that would require showing that $0 \bmod{13}$ isn’t a solution, which is trivial.
11 hours ago history edited Penelope CC BY-SA 4.0
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11 hours ago history asked Penelope CC BY-SA 4.0