jmc

To be divisible by 6, a number must have its digits add up to a multiple of 3, and be even. Therefore, for the hundreds place, the possible digits are $\{5,6,7,8,9\}$, for the tens place the possible digits also are $\{5,6,7,8,9\}$, and for the ones digit, you can only choose from $\{6,8\}$.

First, let us choose 6 for the ones place. The other two digits must add up to a multiple of 3, making a total of 8 pairs that satisfy that condition: $$\{5,7\},\{6,6\},\{6,9\},\{7,5\},\{7,8\},\{8,7\},\{9,6\},\{9,9\}.$$

Next, let us choose 8 for the ones place. The other two digits must be congruent to 1 mod 3, making a total of 8 pairs that satisfy that condition: $$\{5,5\},\{5,8\},\{6,7\},\{7,6\},\{7,9\},\{8,5\},\{8,8\},\{9,7\}.$$

This makes a total of $\boxed{16}$ numbers.