jmc

Adding $-13$ to both sides of $x+13\equiv 55(\mathrm{mod}34)$ gives $x\equiv 55-13(\mathrm{mod}34)$. We find $55-13=42\equiv 8(\mathrm{mod}34)$, so $x\equiv 8(\mathrm{mod}34)$. Thus the smallest positive integer that satisfies the given congruence is $x=8$, and all the other solutions can be obtained by a multiple of 34 to 8. The next three such integers are 42, 76, and 110. Since 110 is greater than 100, there are $\boxed{3}$ integers less than 100 that satisfy the congruence $x+13\equiv 55(\mathrm{mod}34)$.