Difference between revisions of "1996 AHSME Problems/Problem 3"
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==Solution== | ==Solution== | ||
− | The numerator is <math>(3!)! = 6!. | + | The numerator is <math>(3!)! = 6!</math>. |
− | The denominator is < | + | The denominator is <math>3! = 6</math>. |
− | Using the property that < | + | 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}</math>. |
==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 20:11, 18 August 2011
Problem
Solution
The numerator is .
The denominator is .
Using the property that in the numerator, the sixes cancel, leaving , which is answer .
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 |