Difference between revisions of "1996 AHSME Problems/Problem 3"
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| + | ==Problem== | ||
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| + | <math> \frac{(3!)!}{3!}= </math> | ||
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| + | <math> \text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 6\qquad\text{(D)}\ 40\qquad\text{(E)}\ 120 </math> | ||
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| + | ==Solution== | ||
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| + | The numerator is <math>(3!)! = 6!. | ||
| + | |||
| + | The denominator is </math>3! = 6<math>. | ||
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| + | Using the property that </math>6! = 6 \cdot 5!<math> in the numerator, the sixes cancel, leaving </math>5! = 120<math>, which is answer </math>\boxed{E}$. | ||
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==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=2|num-a=4}} | {{AHSME box|year=1996|num-b=2|num-a=4}} | ||
Revision as of 19:11, 18 August 2011
Problem
Solution
The numerator is $(3!)! = 6!.
The denominator is$ (Error compiling LaTeX. Unknown error_msg)3! = 6$.
Using the property that$ (Error compiling LaTeX. Unknown error_msg)6! = 6 \cdot 5!
5! = 120
\boxed{E}$.
See also
| 1996 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
