jmc

With a few initial observations, we realize that the hundreds digit cannot be greater than 1, because the digits 5 or 6 are needed to make a 3 digit integer but will not fit with the larger hundreds digit. Clearly one of the digits is 5, to contribute to the sum $120=5!$. Then since we have the hundreds digit of 1, $1!=1$, we need a middle digit. After testing a few, 4 works, since $145=1!+4!+5!=1+24+120=\boxed{145}$.