Difference between revisions of "2004 AMC 12A Problems/Problem 25"
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Since the only factor in the numerator that goes into <math>98</math> is <math>2</math>, <math>n</math> is minimized. Therefore the answer is <math>13 \cdot 37 \cdot 2=962 \Rightarrow \text {(E)}</math>. | Since the only factor in the numerator that goes into <math>98</math> is <math>2</math>, <math>n</math> is minimized. Therefore the answer is <math>13 \cdot 37 \cdot 2=962 \Rightarrow \text {(E)}</math>. | ||
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+ | ==Solution 2== | ||
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+ | Note that<cmath>0.\overline{133}_n = \frac{n^2+3n+3}{n^3-1},</cmath>by geometric series. | ||
+ | Thus, we're aiming to find the value of<cmath>\prod_{k=4}^{99} \frac{k^2+3k+3}{k^3 - 1}.</cmath>Expanding the product out, this is equivalent to | ||
+ | <cmath>\frac{4^2+3(4)+3}{4^3 - 1} \cdot \frac{5^2+3(5)+3}{5^3 - 1} \cdot \frac{6^2+3(6)+3}{6^3 - 1} \cdot ... \cdot \frac{99^2+3(99)+3}{99^3 - 1}.</cmath>Note that the numerator of the <math>a</math>th fraction and the denominator of the <math>a+1</math>th fraction for <math>1 \leq a \leq 95</math> cancel out to be <math>\frac{1}{a+3},</math> since<cmath>\frac{k^2 + 3k + 3}{(k+1)^3 - 1} = \frac{k^2 + 3k + 3}{k^3 + 3k^2 + 3k} = \frac{1}{k},</cmath>by the binomial theorem on the the denominator of the aforementioned. Since this forms a telescoping series, our product is now equivalent to<cmath>\frac{99^2 + 3(99) + 3}{4^3 - 1} \cdot \prod_{k=4}^{98} \frac{1}{k}, </cmath>which, after simplification gives <math>\frac{6}{98!} \cdot \frac{10101}{63} = \frac{962}{98!},</math> giving an answer of <math>\boxed{962}.</math> | ||
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+ | -fidgetboss_4000 | ||
==See Also== | ==See Also== |
Revision as of 15:59, 18 March 2020
Contents
Problem
For each integer , let denote the base- number . The product can be expressed as , where and are positive integers and is as small as possible. What is the value of ?
Solution
This is an infinite geometric series with common ratio and initial term , so .
Alternatively, we could have used the algebraic manipulation for repeating decimals,
Some factors cancel, (after all, isn't one of the answer choices)
Since the only factor in the numerator that goes into is , is minimized. Therefore the answer is .
Solution 2
Note thatby geometric series. Thus, we're aiming to find the value ofExpanding the product out, this is equivalent to Note that the numerator of the th fraction and the denominator of the th fraction for cancel out to be sinceby the binomial theorem on the the denominator of the aforementioned. Since this forms a telescoping series, our product is now equivalent towhich, after simplification gives giving an answer of
-fidgetboss_4000
See Also
2004 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Question |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.