Rephrase the first part in terms of divisibilities

In general, if a number $A$ divides another number $B > A$ , then all factors of $A$ will also divide $B$ .

For instance, $10$ divides $20$ , so $1$ , $2$ , and $5$ must also divide $20$ .

In this problem,

$\displaystyle \begin{aligned} 2 & \mid N \\ 3 & \mid N \\ 4 & \mid N \end{aligned}$

Thus, $2 \times 3 = \boxed{6}$ , $2 \times 4 = 8$ , $3 \times 4 = 12$ , and $2 \times 3 \times 4 = 24$ must also divide $N$ .

Note that neither $5$ nor $7$ can divide $N$ because no cyclic product of $2$ , $3$ , and $4$ produces $5$ or $7$ . Any prime $p$ will only divide a number $N > p$ if multiples of $p$ also divide $N$ .

$\text{lcm}(2,3,4) = 12,$ which is divisible by 6.

If $N$ is divisible by 2,3 and 4, it must be divisible by 12 and 6.

Note: 12 is not divisible by 5 and 7. Hence these cannot be the answer.