Difference between revisions of "2019 AMC 10A Problems/Problem 18"
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==Solution== | ==Solution== | ||
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+ | We can expand the fraction <math>0.\overline{23}_k</math> as follows: <math>0.\overline{23}_k = 2\cdot k^{-1} + 3 \cdot k^{-2} + 2 \cdot k^{-3} + 3 \cdot k^{-4} + ...</math>. Notice that this is equivalent to | ||
+ | <cmath>2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + ... )</cmath> | ||
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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> | ||
==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
By summing the infinite series and simplifying, we have . Solving this quadratic equation or testing the answer choices yields the answer
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.