AMC 10 · 2014 · #8

Grade 8 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

perfect-squaresfactorialprime-factorization convert-to-algebrapattern-recognition ↑ Prerequisites: factorialperfect-squares

📏 Medium solution 💡 2 insights

Problem

Each answer choice has the form (two consecutive factorials multiplied together) divided by 2: the five choices are 14!·15!/2, 15!·16!/2, 16!·17!/2, 17!·18!/2, and 18!·19!/2. A perfect square is an integer equal to some whole number times itself. Which one of the five choices is a perfect square?

Pick an answer.

(A)

$\dfrac{14!15!}2$

(B)

$\dfrac{15!16!}2$

(C)

$\dfrac{16!17!}2$

(D)

$\dfrac{17!18!}2$

(E)

$\dfrac{18!19!}2$

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

Testing five giant factorial expressions one by one is hopeless — 18! alone has fifteen digits. Tool #5 spots that all five choices are the same object (n! (n+1)!)/2 with n running 14,15,16,17,18. Tool #4 then names that index n and rewrites the shared form so the perfect-square question collapses to a tiny condition on n. Because (n+1)! = (n+1) · n!, the product n! (n+1)! becomes (n!)²(n+1) — a perfect square times a leftover factor. The whole thing is a perfect square exactly when that leftover, (n+1)/2, is itself a perfect square. Tool #3 finishes it: plug each choice's n into that one small test and eliminate until one survives — no factorial ever has to be computed.

1STEP 1

Name the shared form

All five choices are one shape with a changing index — call the smaller factorial n, so the choices are just n = 14, 15, 16, 17, 18.

(n! (n+1)!)/2 for n = 14, 15, 16, 17, 18

2STEP 2

Pull out a perfect square

Since (n+1)! = (n+1)·n!, the product becomes (n!)²(n+1); halving leaves the square (n!)² times (n+1)/2.

(n! (n+1)!)/2 = (n! (n+1) n!)/2 = (n!)²·(n+1)/2

3STEP 3

Reduce to one small condition

A square times something is a square only if that something is too — so everything hinges on (n+1)/2 being a perfect square.

(n!)²·(n+1)/2 is a perfect square⇔ (n+1)/2 is a perfect square

4STEP 4

Test each choice

Run (n+1)/2 for each n: 15/2, 17/2, 19/2 are not integers and 16/2 = 8 is no square, but 18/2 = 9 is.

n=14:15/2 (not integer); n=15:16/2=8; n=16:17/2 (not integer); n=17:18/2=9=3²; n=18:19/2 (not integer)

5STEP 5

Confirm the winner

Only n = 17 makes (n+1)/2 a square: 9 = 3². Rebuilding gives 17!·18!/2 = (17!)²·9 = (3·17!)² → (D).

17! 18!/2 = (17!)² · 9 = (3 · 17!)² → (D)

Answer

17!18!/2

The test says a choice works only when (n+1)/2 is a perfect square, which forces n+1 to be twice a perfect square. Among 15, 16, 17, 18, 19, the even ones are 16 = 2 · 8 and 18 = 2 · 9; of those only 18 hands back a perfect square (9), so exactly one choice qualifies — as an AMC answer should. Checking (D) directly, 17! 18!/2 = (3 · 17!)², a genuine square. The near-misses confirm it: (B) gives 8 (between 2² and 3²), and (A), (C), (E) are not even integers after halving. The answer is (D), matching the key.

💡Key takeaway

A product of consecutive factorials hides a perfect square (n!)² inside it — so all that matters is whether the leftover piece (n+1)/2 is a perfect square too.

Name the shared form
Pull out a perfect square
Reduce to one small condition
Test each choice
Confirm the winner

Name the shared form

Pull out a perfect square

Reduce to one small condition

Test each choice

Confirm the winner

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