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

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 $x = -4, -4*91, -4+(2*91),..$

• 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 $\gcd(21,91)$. – barak manos Nov 20 '16 at 17:36
• Hint $\$ Consider $\ 21x = 1 + 91n\$ modulo $\,7\, [= \gcd(21,91)]\ \$ – Bill Dubuque Nov 20 '16 at 17:40
• – lab bhattacharjee Nov 21 '16 at 17:33

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
$$21x \equiv 1 \mod 7 \implies 0 \equiv 1 \mod 7$$
$21$ is not invertible modulo $91$ since $13\cdot 21=3\cdot 91 \equiv 0\mod 91$.