jmc

Reading all congruences $(\mathrm{mod}42)$, we have $$\begin{array}{rl}a-b& \equiv 18-73\\ & \equiv -55\\ & \equiv -55+42+42\\ & \equiv 29(\mathrm{mod}42).\end{array}$$That's great, except we want to find $n$ with $100\le n<142$. Therefore we should add copies of 42 until we get into this range: $$29\equiv 29+42\equiv 71(\mathrm{mod}42).$$That's not large enough. $$71\equiv 71+42\equiv 113(\mathrm{mod}42).$$That is in our range, so $n=\boxed{113}$.