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2014 AMC 10A Problems/Problem 8: Difference between revisions

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<math>\dfrac{n!(n+1)!}{2}=\dfrac{(n!)^2\cdot(n+1)}{2}=(n!)^2\cdot\dfrac{n+1}{2}</math>. Therefore, the product will only be a perfect square if the second term is a perfect square.
<math>\dfrac{n!(n+1)!}{2}=\dfrac{(n!)^2\cdot(n+1)}{2}=(n!)^2\cdot\dfrac{n+1}{2}</math>. Therefore, the product will only be a perfect square if the second term is a perfect square.
The only answer for which the previous is true is <math>\dfrac{17!18!}{2}=(17!)^2\cdot9</math>.
The only answer for which the previous is true is <math>\dfrac{17!18!}{2}=(17!)^2\cdot9</math>.
==Solution 2==
Notice that <math>17!18!=17!(17!\times 18)=(17!)^2\times 18</math>. So <math>\dfrac{(17!)^2\times 18}{2}=(17!)^2\times 9=(17!)^2\times 3^2</math>. Therefore, it is a perfect square. None of the other choices can be factored this way.


==See Also==
==See Also==

Revision as of 11:12, 9 February 2014

Problem

Which of the following number is a perfect square?

$\textbf{(A)}\ \dfrac{14!15!}2\qquad\textbf{(B)}\ \dfrac{15!16!}2\qquad\textbf{(C)}\ \dfrac{16!17!}2\qquad\textbf{(D)}\ \dfrac{17!18!}2\qquad\textbf{(E)}\ \dfrac{18!19!}2$

Solution 1

Note that $\dfrac{n!(n+1)!}{2}=\dfrac{(n!)^2\cdot(n+1)}{2}=(n!)^2\cdot\dfrac{n+1}{2}$. Therefore, the product will only be a perfect square if the second term is a perfect square. The only answer for which the previous is true is $\dfrac{17!18!}{2}=(17!)^2\cdot9$.

See Also

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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