jmc

Let $a$ be the least number of cookies Mohan could have. From the given information, we know that $$\begin{array}{rl}a& \equiv 3(\mathrm{mod}4)\\ a& \equiv 2(\mathrm{mod}5)\\ a& \equiv 4(\mathrm{mod}7)\end{array}$$ Congruence $(1)$ means that there exists a non-negative integer $m$ such that $a=3+4m$. Substituting this into $(2)$ yields $$3+4m\equiv 2(\mathrm{mod}5)⟹m\equiv 1(\mathrm{mod}5).$$ So there exists a non-negative integer $n$ such that $m=1+5n$. Substituting $a=3+4m$ into $(3)$ yields $$3+4m\equiv 4(\mathrm{mod}7)⟹m\equiv 2(\mathrm{mod}7).$$ Substituting $m=1+5n$ into this gives $$1+5n\equiv 2(\mathrm{mod}7)⟹n\equiv 3(\mathrm{mod}7).$$ The least $n$ such that $n\equiv 3(\mathrm{mod}7)$ is $n=3$. Since $$a=3+4m=3+4(1+5n)=7+20n,$$we have$$n\ge 3⟹a=7+20n\ge 67.$$ Since $67$ satisfies the three congruences, $a=\boxed{67}$.