2004 AMC 10B Problem 6

Peel the larger factorial down to the smaller one. For any $n$,
$$n! \cdot (n+1)! = n! \cdot (n+1) \cdot n! = (n!)^2 \cdot (n+1).$$
Since $(n!)^2$ is already a square, the product is a square exactly when the leftover factor $n + 1$ is a square.

• (A): $(98!)^2 \cdot 99$; $99$ is not a square.
• (C): $(99!)^2 \cdot 100 = (10 \cdot 99!)^2$; a square.
• (E): $(100!)^2 \cdot 101$; $101$ is not a square.

For the choices two apart: (B) $= 98! \cdot 100! = (98!)^2 \cdot 99 \cdot 100$, and $9900$ is not a square because the factor $99 = 9 \cdot 11$ contributes an unpaired $11$. (D) $= (99!)^2 \cdot 100 \cdot 101$, and $101$ is prime, so it is not a square either.

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

Factorials with nearby indices share almost all their factors, so write the bigger one as the smaller one times a few explicit integers. The square part can be ignored; only the small leftover decides the question. A product is a perfect square if and only if every prime appears to an even power, which is why a lone $11$ or $101$ rules a choice out.

• Trying to reason about the huge numbers directly, or guessing that 100! must be involved in a square without checking what factor is left over.
• Believing that $98! \cdot 100!$ is a square because the indices differ by an even number; the leftover $99 \cdot 100$ still contains an unpaired prime.