A whole number is divisible by 4 and is also divisible by 6. Which of these must it be divisible by as well?

Being a multiple of both 4 and 6 makes a number a multiple of their LCM — not of their product. Since $4=2^2$ and $6=2\times 3$, the LCM is $2^2\times 3=12$. The trap is $24=4\times 6$: multiplying the two divisors is only valid when they are co-prime, and here they share a factor of 2. The number 12 itself settles it — it is divisible by 4 and by 6, yet not by 24, 16 or 36.