Difference between revisions of "2008 AMC 12A Problems/Problem 4"

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{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #4]] and [[2008 AMC 10A Problems/Problem 5|2008 AMC 10A #5]]}}
 
{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #4]] and [[2008 AMC 10A Problems/Problem 5|2008 AMC 10A #5]]}}
==Problem==
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== Problem ==
 
Which of the following is equal to the [[product]]
 
Which of the following is equal to the [[product]]
 
<cmath>\frac{8}{4}\cdot\frac{12}{8}\cdot\frac{16}{12}\cdot\cdots\cdot\frac{4n+4}{4n}\cdot\cdots\cdot\frac{2008}{2004}?</cmath>
 
<cmath>\frac{8}{4}\cdot\frac{12}{8}\cdot\frac{16}{12}\cdot\cdots\cdot\frac{4n+4}{4n}\cdot\cdots\cdot\frac{2008}{2004}?</cmath>
  
<math>\mathrm{(A)}\ 251\qquad\mathrm{(B)}\ 502\qquad\mathrm{(C)}\ 1004\qquad\mathrm{(D)}\ 2008\qquad\mathrm{(E)}\ 4016</math>
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<math>\textbf{(A)}\ 251\qquad\textbf{(B)}\ 502\qquad\textbf{(C)}\ 1004\qquad\textbf{(D)}\ 2008\qquad\textbf{(E)}\ 4016</math>
  
 
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Notice that everything cancels out except for <math>2008</math> in the numerator and <math>4</math> in the denominator.
 
Notice that everything cancels out except for <math>2008</math> in the numerator and <math>4</math> in the denominator.
  
Thus, the product is <math>\frac{2008}{4}=502</math>, and the answer is <math>\mathrm{(B)}</math>.  
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Thus, the product is <math>\frac{2008}{4}=502</math>, and the answer is <math>\textbf{(B)}</math>.  
  
 
==See Also==  
 
==See Also==  
 
{{AMC12 box|year=2008|ab=A|num-b=3|num-a=5}}
 
{{AMC12 box|year=2008|ab=A|num-b=3|num-a=5}}
 
{{AMC10 box|year=2008|ab=A|num-b=4|num-a=6}}
 
{{AMC10 box|year=2008|ab=A|num-b=4|num-a=6}}
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[[Category:Introductory Algebra Problems]]

Revision as of 01:38, 26 February 2010

The following problem is from both the 2008 AMC 12A #4 and 2008 AMC 10A #5, so both problems redirect to this page.

Problem

Which of the following is equal to the product \[\frac{8}{4}\cdot\frac{12}{8}\cdot\frac{16}{12}\cdot\cdots\cdot\frac{4n+4}{4n}\cdot\cdots\cdot\frac{2008}{2004}?\]

$\textbf{(A)}\ 251\qquad\textbf{(B)}\ 502\qquad\textbf{(C)}\ 1004\qquad\textbf{(D)}\ 2008\qquad\textbf{(E)}\ 4016$

Solution

Solution 1

$\frac {8}{4}\cdot\frac {12}{8}\cdot\frac {16}{12}\cdots\frac {4n + 4}{4n}\cdots\frac {2008}{2004} = \frac {1}{4}\cdot\left(\frac {8}{8}\cdot\frac {12}{12}\cdots\frac {4n}{4n}\cdots\frac {2004}{2004}\right)\cdot 2008 = \frac{2008}{4} =$ $502 \Rightarrow B$.

Solution 2

Notice that everything cancels out except for $2008$ in the numerator and $4$ in the denominator.

Thus, the product is $\frac{2008}{4}=502$, and the answer is $\textbf{(B)}$.

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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 12 Problems and Solutions
2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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