jmc

For a number to be divisible by $3$, its digits must add up to be a number divisible by $3$. We can easily find that three combinations work: $2+7=9$, $5+7=12$, and $3+9=12$. Since the unit and tens digits are interchangeable (the digits will still add up to a multiple of 3), there are $3\cdot 2=\boxed{6}$ possible numbers.