2014 AMC 10A Problem 8

Below is the professionally curated solution for Problem 8 of the 2014 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 10A solutions, or check the answer key.

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Concepts:factorialperfect squareprime factorization

Difficulty rating: 1420

8.

Which of the following numbers is a perfect square?

14!15!2\dfrac{14!15!}2

15!16!2\dfrac{15!16!}2

16!17!2\dfrac{16!17!}2

17!18!2\dfrac{17!18!}2

18!19!2\dfrac{18!19!}2

Solution:

Note that all of these answer choices are of the form n!(n+1)!2=(n!)2(n+1)2. \dfrac{n!(n + 1)!}{2} = \dfrac{(n!)^2(n + 1)}{2}. We have that (n!)2(n!)^2 is square, so we need n+12\dfrac{n + 1}{2} to be square as well.

This means that n+1n + 1 must be twice a perfect square. The only choice we have is n+1=18,n + 1 = 18, which gives us n=17.n = 17.

Thus, D is the correct answer.

Problem 8 in Other Years