2014 AMC 10A Problem 8

The numbers are far too big to compute, so look for a square hidden inside them.

1. Take choice (D). Notice that $18! = 18 \cdot 17!$, because $18!$ is $17!$ with one more factor, $18$.
2. So
$$\frac{17!\,18!}{2} = \frac{17! \cdot 17! \cdot 18}{2} = (17!)^2 \cdot 9.$$
3. $(17!)^2$ is a perfect square, and $9 = 3^2$ is a perfect square. A square times a square is a square, so this number equals $(17! \cdot 3)^2$.
4. Check that the others fail. Each choice has the form $\frac{n!\,(n+1)!}{2} = (n!)^2 \cdot \frac{n+1}{2}$, so only the leftover $\frac{n+1}{2}$ matters:

The answer is $\boxed{\textbf{(D)}\ \dfrac{17!18!}2}$.

Two factorials next to each other share almost all their factors, so their product is a square times one extra number. Pull out the part that is clearly a square, and the question shrinks to whether the small leftover is a square. Look for a hidden square like this before counting prime factors.

Every choice has the form $\dfrac{n!\,(n+1)!}{2}$. Since $(n+1)! = (n+1)\cdot n!$,
$$\frac{n!\,(n+1)!}{2} = (n!)^2 \cdot \frac{n+1}{2}.$$
The factor $(n!)^2$ is already a perfect square, so the whole number is a perfect square exactly when $\frac{n+1}{2}$ is a perfect square.

Check the five values of $n$:
- $n = 14$: $\frac{15}{2}$, not an integer.
- $n = 15$: $\frac{16}{2} = 8$, not a square.
- $n = 16$: $\frac{17}{2}$, not an integer.
- $n = 17$: $\frac{18}{2} = 9 = 3^2$. Yes.
- $n = 18$: $\frac{19}{2}$, not an integer.

The answer is $\boxed{\textbf{(D)}\ \dfrac{17!18!}2}$.