jmc

We have to consider 2 cases:

When the hundreds digit is 3, we want the sum of the tens and unit digit equal to 13. We have $4+9=5+8=6+7=13,$ which yield a total of 6 choices (two for each pair that adds to 13).

When the hundreds digit is 4, we want the sum of the tens and unit digit equal to 12. We have $3+9=4+8=5+7=6+6=12.$ The first three pairs give us 2 solutions, but the last only gives 1, so we have a total of 7 choices.

Therefore there are total of $6+7=\boxed{13}$ integers.