jmc

Let $a$ be the desired number. The given system of congruences is $$\begin{array}{r}a\equiv 1(\mathrm{mod}3)\\ a\equiv 1(\mathrm{mod}4)\\ a\equiv 2(\mathrm{mod}5)\end{array}$$ Since $\gcd (3,4)=1$, $(1)$ and $(2)$ together imply that $a\equiv 1(\mathrm{mod}12)$. So there exists a non-negative integer $n$ such that $a=1+12n$. Substituting this into $(3)$ yields $$1+12n\equiv 2(\mathrm{mod}5),$$ $$⟹n\equiv 3(\mathrm{mod}5).$$ So $n$ has a lower bound of $3$. Then $$n\ge 3,$$ $$⟹a=1+12n\ge 37.$$ Since $37$ satisfies all three congruences, $a=\boxed{37}$.