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

(Solution 2)
(Solution 3 (extremely rigorous))
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==Solution 3 (extremely rigorous)==
 
==Solution 3 (extremely rigorous)==
 
Plug in the values of k and bash.  This gives us <math>\boxed{D}</math>.
 
Plug in the values of k and bash.  This gives us <math>\boxed{D}</math>.
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==Solution 4==
 +
Similar to Solution 1, we arrive at <math>\frac{2k+3}{k^2-1}=\frac{7}{51}</math>. We can rewrite this as <math>\frac{2k+3}{(k-1)(k+1)}=\frac{7}{51}=\frac{7}{3\cdot 17}</math>. Notice that <math>2k+3=2(k+1)+1=2(k-1)+5</math>. As <math>17</math> is a prime, we have that one of <math>k-1</math> and <math>k+1</math> is divisible by <math>17</math>. Looking at the answer choices, this gives <math>\boxed{\textbf{(D) }16}</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 23:45, 10 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 yields $k=16$, so the answer is $\boxed{D}$

-eric2020

Solution 3 (extremely rigorous)

Plug in the values of k and bash. This gives us $\boxed{D}$.

Solution 4

Similar to Solution 1, we arrive at $\frac{2k+3}{k^2-1}=\frac{7}{51}$. We can rewrite this as $\frac{2k+3}{(k-1)(k+1)}=\frac{7}{51}=\frac{7}{3\cdot 17}$. Notice that $2k+3=2(k+1)+1=2(k-1)+5$. As $17$ is a prime, we have that one of $k-1$ and $k+1$ is divisible by $17$. Looking at the answer choices, this gives $\boxed{\textbf{(D) }16}$.

Video Solution

For those who want a video solution : https://www.youtube.com/watch?v=DFfRJolhwN0

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

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