, 2024

Solution:

The prime divisors of $72$ are $2$ and $3$, and we note that $72 = 2^{3} \cdot 3^{2}$. Since we need at least one factor of $2$ and one factor of $3$, we just need to check the multiples of $6$ from $72$ onwards.

$13 \cdot 6$, $14 \cdot 6$, and $15 \cdot 6$ have prime factors of $13$, $7$, and $5$ respectively; thus we get $96 = 2^{5} \cdot 3 = 16 \cdot 6$ as the smallest number that satisfies the condition given in the problem.