jmc

Note that it is possible that $a=1$ and $b=2$, since $3\cdot 2=8-2\cdot 1$. Then $2b+12=16$. Since $3,$ $5,$ and $6,$ are not factors of $16$, it is not true that these numbers must be divisors of $2b+12.$

It only remains to check whether $1$, $2$, and $4$ must be divisors of $2b+12$. The distributive property gives us $$8-2a=2\cdot 4-2a=2(4-a),$$so $2$ is a factor of $3b$. Note that $$b=3b-2b,$$so since $2$ is a factor of $3b$ and $2$ is a factor of $2b,$ $2$ must be a factor of $b.$ So we can say $b=2n$ for some integer $n$. Substituting gives us $$\begin{array}{rl}2b+12& =2(2n)+12\\ & =4n+4\cdot 3\\ & =4(n+3),\end{array}$$so $4$ is a factor of $2b+12$. Since $1,$ $2,$ and $4$ are factors of $4$ and $4$ is a factor of $2b+12$, it must be true that $1,$ $2,$ and $4$ are factors of $2b+12$. So our final answer is $\boxed{3}.$