2014 AMC 10A Problem 8

P1-10~2 minPrint

Number Properties Divisibility & Factors Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Use the answer choices (mod checks, size, form) to eliminate or select

Which of the following numbers is a perfect square?

Show solution without answering

Key insight

The two factorials share a big common part, so each choice is a perfect square times (n+1)/2. Only 18/2 = 9 is a perfect square.

Solution

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:

choice	$n$	$\frac{n+1}{2}$	perfect square?
(A)	$14$	$\frac{15}{2}$	no, not a whole number
(B)	$15$	$8$	no
(C)	$16$	$\frac{17}{2}$	no, not a whole number
(D)	$17$	$9$	yes
(E)	$18$	$\frac{19}{2}$	no, not a whole number

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 11: Special Factorizations
Introduction to Algebra, Chapter 20: Special Functions (§20.3)
Introduction to Number Theory, Chapter 5: Divisor Problems (§5.2, §5.3)