Difference between revisions of "2008 AMC 12A Problems/Problem 4"
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− | ==Problem== | + | {{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #4]] and [[2008 AMC 10A Problems/Problem 5|2008 AMC 10A #5]]}} |
+ | |||
+ | == 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>\ | + | <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>\ | + | 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}} | ||
+ | |||
+ | |||
+ | [[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
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.