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> | ||
| − | == Solution == | + | == Solution 1== |
Note that | 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. | <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>. | The only answer for which the previous is true is <math>\dfrac{17!18!}{2}=(17!)^2*9</math>. | ||
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| + | ==Solution 2== | ||
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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</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
Contents
Problem
Which of the following number is a perfect square?
Solution 1
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 
.
Solution 2
Notice that 
. So 
. Therefore, it is a perfect square. None of the other choices can be factored this way.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
