Difference between revisions of "2014 AMC 10A Problems/Problem 8"
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<math> \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 </math> | <math> \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 </math> | ||
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| + | == Solution == | ||
| + | |||
| + | Note that | ||
| + | <math>\dfrac{n!(n+1)!}{2}=\dfrac{(n!)^2*(n+1)}{2}=(n!)^2*\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*9</math>. | ||
Revision as of 21:50, 6 February 2014
Problem
Which of the following number is a perfect square?
Solution
Note that
. 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 
.
