Difference between revisions of "2019 AMC 10A Problems/Problem 18"
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By summing the infinite series and simplifying, we have <math>\frac{2k+3}{k^2-1} = \frac{7}{51}</math>. Solving this quadratic equation or testing the answer choices yields the answer <math>\boxed{k=16}.</math> | By summing the infinite series and simplifying, we have <math>\frac{2k+3}{k^2-1} = \frac{7}{51}</math>. Solving this quadratic equation or testing the answer choices yields the answer <math>\boxed{k=16}.</math> | ||
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| + | -- OmicronGamma | ||
==See Also== | ==See Also== | ||
Revision as of 18:13, 9 February 2019
- The following problem is from both the 2019 AMC 10A #18 and 2019 AMC 12A #11, so both problems redirect to this page.
Problem
For some positive integer 
, the repeating base- representation of the (base-ten) fraction 
is 
. What is ?
Solution
We can expand the fraction 
as follows: 
. Notice that this is equivalent to
![\[2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + ... )\]](http://latex.artofproblemsolving.com/1/8/b/18bcad6f00558b8c5122c7996d6f0683037c748b.png)
By summing the infinite series and simplifying, we have 
. Solving this quadratic equation or testing the answer choices yields the answer 
-- OmicronGamma
See Also
| 2019 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 | ||
| 2019 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 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. 
