2024 AMC 10B Problem 2

These numbers are huge, so do not multiply them out. Instead, look for a piece the two terms share.

1. $10!$ means $10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdots 1$. The part from $7$ down to $1$ is exactly $7!$, so
$$10! = 10 \cdot 9 \cdot 8 \cdot 7!.$$
2. Work out the front part: $10 \cdot 9 \cdot 8 = 720$. So $10! = 720 \cdot 7!$.
3. Now look at $6!$: $6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 720$. So the second term is $7! \cdot 6! = 7! \cdot 720$.
4. The two terms are the same number, so
$$10! - 7! \cdot 6! = 720 \cdot 7! - 720 \cdot 7! = 0.$$

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

Factorials hide a lot of shared factors. Pulling out the biggest factorial that both terms contain ($7!$ here) leaves small numbers you can compare easily. The whole problem rests on the surprise that $10 \cdot 9 \cdot 8$ and $6!$ are both $720$.

Both $10!$ and $7! \cdot 6!$ have $7$ as a factor, so their difference is also a multiple of $7$. Check the choices: $120$, $600$ and $720$ are not multiples of $7$, and neither is $-120$. Only $0$ is, so the answer is (B).

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 $\boxed{\textbf{(B)}\ 0}$.