Difference between revisions of "1996 AHSME Problems/Problem 3"
Talkinaway (talk | contribs) (→See also) |
|||
| (2 intermediate revisions by one other user not shown) | |||
| Line 1: | Line 1: | ||
| + | ==Problem== | ||
| + | |||
| + | <math> \frac{(3!)!}{3!}= </math> | ||
| + | |||
| + | <math> \text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 6\qquad\text{(D)}\ 40\qquad\text{(E)}\ 120 </math> | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | The numerator is <math>(3!)! = 6!</math>. | ||
| + | |||
| + | The denominator is <math>3! = 6</math>. | ||
| + | |||
| + | 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}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:06, 5 July 2013
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
