Difference between revisions of "1996 AHSME Problems/Problem 3"

(Solution)
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==Solution==
 
==Solution==
  
The numerator is <math>(3!)! = 6!.
+
The numerator is <math>(3!)! = 6!</math>.
  
The denominator 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}$.
+
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

$\frac{(3!)!}{3!}=$

$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 6\qquad\text{(D)}\ 40\qquad\text{(E)}\ 120$

Solution

The numerator is $(3!)! = 6!$.

The denominator is $3! = 6$.

Using the property that $6! = 6 \cdot 5!$ in the numerator, the sixes cancel, leaving $5! = 120$, which is answer $\boxed{E}$.

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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