Solve $22x \equiv 5(mod 15)$

I looked at an example of this type, and here's my attempt:

$gcd(22,15)=1$ and $1$ is a divisor of $5$ so solutions exist.

Now $22x \equiv 5(mod 15)$ is the same as solving $22x=5$ in $Z_{15}$ i.e. $7x=5$ in $Z_{15}$

The units in $Z_{15}$ are $1,2,4,7,8,11,13,14$

$7\times 13=91=1$ in $Z_{15}$, so $13$ is the inverse of $7$.

so, $x=13\times 5= 65=5$ in $Z_{15}$

Thus, the solutions are $5+15\mathbb Z$

Am I correct?

1 Answer

You are correct. ${}{}{}{}{}{}$