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

m (Solution 1)
(Improved clarity and formatting)
Line 9: Line 9:
 
==Solution 1==
 
==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  
 
<cmath>2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + ... )</cmath>
 
<cmath>2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + ... )</cmath>
  
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>k = \boxed{\textbf{(D) }16}.</math>
+
By summing the geometric series and simplifying, we have <math>\frac{2k+3}{k^2-1} = \frac{7}{51}</math>. Solving this quadratic equation (or simply testing the answer choices) yields the answer <math>k = \boxed{\textbf{(D) }16}</math>.
  
 
==Solution 2==
 
==Solution 2==
Line 18: Line 18:
 
Let <math>a = 0.2323\dots_k</math>. Therefore, <math>k^2a=23.2323\dots_k</math>.
 
Let <math>a = 0.2323\dots_k</math>. Therefore, <math>k^2a=23.2323\dots_k</math>.
  
From this, we see that <math>k^2a-a=23_k</math>.  
+
From this, we see that <math>k^2a-a=23_k</math>, so <math>a = \frac{23_k}{k^2-1} = \frac{2k+3}{k^2-1} = \frac{7}{51}</math>.
  
Solving for a:
+
Now, similar to in Solution 1, we can either test if <math>2k+3</math> is a multiple of 7 with the answer choices, or actually solve the quadratic, so that the answer is <math>\boxed{\textbf{(D) }16}</math>.
  
<math>a = \frac{23_k}{k^2-1} = \frac{2k+3}{k^2-1} = \frac{7}{51}</math>
+
==Solution 3 (bash)==
 +
We can simply plug in all the answer choices as values of <math>k</math>, and see which one works. After lengthy calculations, this eventually gives us <math>\boxed{\textbf{(D) }16}</math> as the answer.
  
Similar to Solution 1, testing if <math>2k+3</math> is a multiple of 7 with the answer choices or solving the quadratic yields <math>k=16</math>, so the answer is <math>\boxed{D}</math>.
+
==Solution 4==
 +
Just as in Solution 1, we arrive at the equation <math>\frac{2k+3}{k^2-1}=\frac{7}{51}</math>.
  
==Solution 3 (extremely rigorous)==
+
We can now 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 therefore must have that one of <math>k-1</math> and <math>k+1</math> is divisible by <math>17</math>. Now, checking each of the answer choices, this gives <math>\boxed{\textbf{(D) }16}</math>.
Plug in the values of k and bash.  This gives us <math>\boxed{D}</math>.
 
 
 
==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 00:21, 27 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 geometric series and simplifying, we have $\frac{2k+3}{k^2-1} = \frac{7}{51}$. Solving this quadratic equation (or simply testing the answer choices) yields the answer $k = \boxed{\textbf{(D) }16}$.

Solution 2

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

From this, we see that $k^2a-a=23_k$, so $a = \frac{23_k}{k^2-1} = \frac{2k+3}{k^2-1} = \frac{7}{51}$.

Now, similar to in Solution 1, we can either test if $2k+3$ is a multiple of 7 with the answer choices, or actually solve the quadratic, so that the answer is $\boxed{\textbf{(D) }16}$.

Solution 3 (bash)

We can simply plug in all the answer choices as values of $k$, and see which one works. After lengthy calculations, this eventually gives us $\boxed{\textbf{(D) }16}$ as the answer.

Solution 4

Just as in Solution 1, we arrive at the equation $\frac{2k+3}{k^2-1}=\frac{7}{51}$.

We can now 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 therefore must have that one of $k-1$ and $k+1$ is divisible by $17$. Now, checking each of 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png