**Problem 8:**

If $3$ and $4$ individually divides a number, then also divides that number. For example, all of $3, 4$ and $12$ divide $48$. But if a number is divisible by $3$ and $6$ individually it may or may not be divisible by $3 \times 6=18$ . For example, both of $54$ and $60$ are divisible by $3$ and $6$ individually, though only $54$ is divisible by $18$. Can you explain why this happens?