2004 AMC 10B 第 6 题

For $m \lt n,$ write $m! \cdot n!$ as $(m!)^2 \cdot (m+1)(m+2)\cdots n$

The product is a perfect square exactly when $(m+1)(m+2)\cdots n$ is

For $m \lt n,$ $m! \cdot n!$ equals $(m!)^2 \cdot (m+1)(m+2)\cdots n,$ which is a perfect square precisely when $(m+1)\cdots n$ is a perfect square.

For the five choices this leftover factor is $99,$ $99 \cdot 100,$ $100,$ $100 \cdot 101,$ and $101.$ Only $100 = 10^2$ is a perfect square.

Therefore $99! \cdot 100! = (99! \cdot 10)^2$ is the perfect square.

Thus, the correct answer is C.