Difference between revisions of "2014 AMC 10A Problems/Problem 8"
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<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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| + | ==See Also== | ||
| + | |||
| + | {{AMC10 box|year=2014|ab=A|num-b=7|num-a=9}} | ||
| + | {{MAA Notice}} | ||
Revision as of 22:11, 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 
.
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. 
