Difference between revisions of "2008 AMC 12A Problems/Problem 4"
m (Semi-automated contest formatting - script by azjps) |
|||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 1: | Line 1: | ||
{{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 == | == Problem == | ||
Which of the following is equal to the [[product]] | Which of the following is equal to the [[product]] | ||
| Line 6: | Line 7: | ||
<math>\textbf{(A)}\ 251\qquad\textbf{(B)}\ 502\qquad\textbf{(C)}\ 1004\qquad\textbf{(D)}\ 2008\qquad\textbf{(E)}\ 4016</math> | <math>\textbf{(A)}\ 251\qquad\textbf{(B)}\ 502\qquad\textbf{(C)}\ 1004\qquad\textbf{(D)}\ 2008\qquad\textbf{(E)}\ 4016</math> | ||
| − | + | == Solution 1 == | |
| − | |||
| − | |||
<math>\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} =</math> <math>502 \Rightarrow B</math>. | <math>\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} =</math> <math>502 \Rightarrow B</math>. | ||
| − | + | == Solution 2 == | |
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>\textbf{(B)}</math>. | 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}} | ||
| Line 22: | Line 21: | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 03:52, 21 July 2022
- The following problem is from both the 2008 AMC 12A #4 and 2008 AMC 10A #5, so both problems redirect to this page.
Contents
Problem
Which of the following is equal to the product
![\[\frac{8}{4}\cdot\frac{12}{8}\cdot\frac{16}{12}\cdot\cdots\cdot\frac{4n+4}{4n}\cdot\cdots\cdot\frac{2008}{2004}?\]](http://latex.artofproblemsolving.com/6/1/d/61d211dd54e4b681761aa2c6c2fea286d329138e.png)
Solution 1
.
Solution 2
Notice that everything cancels out except for 
in the numerator and 
in the denominator.
Thus, the product is 
, and the answer is 
.
See Also
| 2008 AMC 12A (Problems • Answer Key • Resources) | |
| 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 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
