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 $k$, the repeating base-$k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$. What is $k$?

$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$

Solution

We can expand the fraction $0.\overline{23}_k$ as follows: $0.\overline{23}_k = 2\cdot k^{-1} + 3 \cdot k^{-2} + 2 \cdot k^{-3} + 3 \cdot k^{-4} + ...$. Notice that this is equivalent to \[2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + ... )\]

By summing the infinite series and simplifying, we have $\frac{2k+3}{k^2-1} = \frac{7}{51}$. Solving this quadratic equation or testing the answer choices yields the answer $\boxed{k=16}.$

-- OmicronGamma

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png

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