Difference between revisions of "2014 AMC 10A Problems/Problem 8"

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==Solution 2==
 
==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</math>. Therefore, it is a perfect square. None of the other choices can be factored this way.
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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 22:12, 6 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 $\dfrac{n!(n+1)!}{2}=\dfrac{(n!)^2*(n+1)}{2}=(n!)^2*\dfrac{n+1}{2}$. 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 $\dfrac{17!18!}{2}=(17!)^2*9$.


Solution 2

Notice that $17!18!=17!(17!\times 18)=(17!)^2\times 18$. So $\dfrac{(17!)^2\times 18}{2}=(17!)^2\times 9=(17!)^2\times 3^2$. Therefore, it is a perfect square. None of the other choices can be factored this way.

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
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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