jmc

The numbers that leave a remainder of 2 when divided by 4 and 5 are 22, 42, 62, and so on. Checking these numbers for a remainder of 2 when divided by both 3 and 6 yields $\boxed{62}$ as the smallest.

We also might have noted that the desired number is 2 greater than a number that is a multiple of 3, 4, 5, and 6. Therefore, it is 2 greater than the least common multiple of 3, 4, 5, and 6. The least common multiple of 3, 4, 5, and 6 is $2^2\cdot 3\cdot 5=60$, so the smallest number that fits the problem is $60+2=\boxed{62}$.