2024 AMC 10B Problem 2

Write $10!$ in terms of $7!$:
$$10! = 10 \cdot 9 \cdot 8 \cdot 7! = 720 \cdot 7!.$$
But $6! = 720$ as well, so $7! \cdot 6! = 720 \cdot 7!$ too. The two terms are identical and their difference is
$$720 \cdot 7! - 720 \cdot 7! = 0.$$

The answer is $\boxed{\textbf{(B)}\ 0}$.

Factorials share huge common factors, so never expand them; peel off the largest factorial common to both terms and compare what remains. The coincidence $10 \cdot 9 \cdot 8 = 6!$ is the whole problem. As a sanity check, both terms are multiples of $7$, so the answer must be too, and $0$ is the only choice divisible by $7$.

Both $10!$ and $7! \cdot 6!$ are divisible by $7$, so their difference is a multiple of $7$. Among the choices only $0$ is divisible by $7$.

• Trying to multiply out 10! and 7! * 6! as decimal numbers and making an arithmetic slip somewhere in the seven-digit products.
• Misreading $7! \cdot 6!$ as $7! + 6!$ or as $13!$, which gives a huge nonzero result.