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Difference between revisions of "2019 AMC 10A Problems/Problem 18"

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<math>\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17</math>
 
<math>\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17</math>
  
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==Solution==
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==Solution 1==
  
 
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  
 
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  
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-- OmicronGamma
 
-- OmicronGamma
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==Solution 2==
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Let <math>a = 0.2323\dots_k</math>. Therefore, <math>k^2a=23.2323\dots_k</math>.
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From this, we see that <math>k^2a-a=23_k</math>.
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Solving for a:
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<math>a = \frac{23_k}{k^2-1} = \frac{2k+3}{k^2-1} = \frac{7}{51}</math>
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Similar to Solution 1, testing if <math>2k+3</math> is a multiple of 7 with the answer choices or solving the quadratic will yield <math>k=16</math>, so the answer is <math>\boxed{D}</math>
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-eric2020
  
 
==See Also==
 
==See Also==

Revision as of 18:24, 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 1

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

Solution 2

Let $a = 0.2323\dots_k$. Therefore, $k^2a=23.2323\dots_k$.

From this, we see that $k^2a-a=23_k$.

Solving for a:

$a = \frac{23_k}{k^2-1} = \frac{2k+3}{k^2-1} = \frac{7}{51}$

Similar to Solution 1, testing if $2k+3$ is a multiple of 7 with the answer choices or solving the quadratic will yield $k=16$, so the answer is $\boxed{D}$

-eric2020

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

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