2004 AMC 10B Problem 6

The numbers are huge, so do not try to compute them. Instead, use the fact that a bigger factorial contains the smaller one.

1. Start with a small example: $4! = 4 \times 3!$, because $4! = 4 \times 3 \times 2 \times 1$. In the same way, $100! = 100 \times 99!$.
2. So choice (C) is
$$99! \cdot 100! = 99! \cdot 100 \cdot 99! = (99!)^2 \cdot 100.$$
3. $(99!)^2$ is a square, and $100 = 10^2$ is a square. A square times a square is a square: this is $(99! \cdot 10)^2$. So (C) works.
4. Check that the others fail the same way, by pulling out a square and looking at what is left:
- (A) $98! \cdot 99! = (98!)^2 \cdot 99$. Leftover $99 = 9 \cdot 11$ has a single $11$, so not a square.
- (B) $98! \cdot 100! = (98!)^2 \cdot 99 \cdot 100$. Leftover $9900 = 100 \cdot 9 \cdot 11$, again a single $11$.
- (D) $99! \cdot 101! = (99!)^2 \cdot 100 \cdot 101$. Leftover has a single $101$, which is prime.
- (E) $100! \cdot 101! = (100!)^2 \cdot 101$. Leftover $101$ is prime, not a square.

The answer is $\boxed{\textbf{(C)}\ 99! \cdot 100!}$.

Factorials that are close together share almost all their factors. Rewrite the bigger one as the smaller one times a few extra numbers, pull out the square, and only the small leftover matters. A number is a perfect square exactly when every prime in it shows up an even number of times, so a lone $11$ or $101$ rules a choice out.