I want to show that in ℤ6, the equation
[2]X = [0]
has a solution, but the equation
[2]X = [5]
does not. However, I am unsure how to find the solution to X, specifically, how to divide residue classes? I know that how addition and multiplication work for residue classes, but not division.
Just list all the $2a$ values.
$2*0 \equiv 0$ and $2*1 \equiv 2$ and $2*2 \equiv 4$ and $2*3 \equiv 0$ and $2*4 \equiv 2$ and $2*5 \equiv 4$.
So $2x \equiv 0,2,4$ all have two solutions and $2x \equiv 1,3,5$ have no solutions.
A bit more sophisticated: $2x \equiv 0 \mod 6$ means $6|2x$ which means there is an integer $k$ so that $2x = 6k$ or $x = 3k$ so we con have $x \equiv 0,3$.
$2x \equiv 6 \mod 6$ means $6|2x - 5$ so there is an integer $k$ so that $2x -5 = 6k$. Or $2x = 6k -5$ so $x = 3k - \frac 52$. $\frac 52$ is not an integer so $x$ is never an integer.
A lot more sophisticated: $mx \equiv a \mod n$ means $n|mx - a$ which means $\frac n{\gcd(n,m)}\mid \frac m{\gcd(n,m)}x - \frac a{\gcd(n,m)}$ so unless $\gcd(n,m)|a$ there will be no solution.
Thus as $\gcd(2,6) = 2$ and $2\not \mid 5$ there can not be any solution to $2x \equiv 5$ as $2$ will always divide $2x$ but will never divide $6k + 5$.
If $\gcd(n,m)|a$ and $\frac a{\gcd(n,m)} = b$ then there is a solution as $\gcd(\frac m{\gcd(n,m)}, \frac n{\gcd(n,m)}) = 1$ and for two number being relatively prime $\frac m{\gcd(n,m)}x - \frac n{\gcd(n,m)}k = b$ will always have a solution by Bezout's lemma.
So $2x \equiv 0 \mod 6$ will have as as $\gcd(2,6) =2$ and $\gcd(2,6)|0$.