2014 AMC 10A Problem 8

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

Strip away the part that is obviously a square and examine what is left. Because $n!$ and $(n+1)!$ share the factor $n!$, their product is a square times $n+1$, which turns an intimidating factorial question into "is $(n+1)/2$ a square?" Look for hidden squares before doing any prime-factor counting.

• Trying to compute or factor the huge factorials instead of pulling out the square factor (n!)^2.
• Checking whether $n+1$ (rather than $\frac{n+1}{2}$) is a square, which selects $n = 15$ and choice (B).