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Solve $21x\equiv 1\pmod {91}$

Well, I'm not sure if there're 7 solutions or none. because:

$gcd(21,91) = 7$, but the condition $ 7 \mid 1 $ is false. so I believe there're no solutions, I also can't divide by 7, because $1/7$ won't be equal to $1$.

If I'm wrong so the solutions would be $x = -4, -4*91, -4+(2*91),..$

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    $\begingroup$ This question is a mess, both in terms of mathematical notations and English grammar, but there are no solutions if that's what you're asking, since for any integer value of $x$, the value of $21x\bmod91$ must be divisible by $\gcd(21,91)$. $\endgroup$ – barak manos Nov 20 '16 at 17:36
  • $\begingroup$ Hint $\ $ Consider $\ 21x = 1 + 91n\ $ modulo $\,7\, [= \gcd(21,91)]\ \ $ $\endgroup$ – Bill Dubuque Nov 20 '16 at 17:40
  • $\begingroup$ See proofwiki.org/wiki/Solution_of_Linear_Congruence $\endgroup$ – lab bhattacharjee Nov 21 '16 at 17:33
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The equation has no solutions, since $21$ and $91$ are divisible by $7$ but $1$ is not divisible by $7$. In other words:

$$21x \equiv 1 \mod 91 \implies 21x \equiv 1 \mod 7$$

since $7$ divides $91$. But

$$21x \equiv 1 \mod 7 \implies 0 \equiv 1 \mod 7$$

so the equation has no solutions.

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$21$ is not invertible modulo $91$ since $13\cdot 21=3\cdot 91 \equiv 0\mod 91$.

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