Difference between revisions of "2008 AMC 10A Problems/Problem 5"
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==Solution== | ==Solution== | ||
| − | {{ | + | 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>. | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=A|num-b=4|num-a=6}} | {{AMC10 box|year=2008|ab=A|num-b=4|num-a=6}} | ||
Revision as of 22:57, 25 April 2008
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
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 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 | ||
