jmc

If the number is divisible by $3$, the sum of the digits must be divisible by $3$. So $a+b$ must be one more than a multiple of $3$, since $5$ is one less than a multiple of $3$. We have several options:

If $a+b=1$, it must be $(1,0)$, one possibility.

For $a+b=4$, $a$ can be $1$ to $4$, for four possibilities.

For $a+b=7$, $a$ can be $1$ to $7$, so seven possibilities.

For $a+b=10$, anything from $(1,9)$ to $(9,1)$ works, so nine possibilities.

For $a+b=13$, the pairs range from $(4,9)$ to $(9,4)$, for six possibilities.

If $a+b=16$, we can have $(7,9)$, $(8,8)$, or $(9,7)$, so three possibilities.

$a+b=19$ and up is not possible.

So the total number is

$$1+4+7+9+6+3=\boxed{30}$$