Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 5)
m (→‎Solution 5)
Line 34: Line 34:
 
==Solution 5==
 
==Solution 5==
  
−
Note that the LHS equals <cmath>\bigg(\frac{2}{k} + \frac{2}{k^3} + \cdots \bigg) + \bigg(\frac{3}{k^2} + \frac{3}{k^4} + \cdots \bigg) = \frac{\frac{2}{k}}{1 - \frac{1}{k^2-1}} + \frac{\frac{3}{k^2}}{1 - \frac{1}{k^2}} = \frac{2k+3}{k^2},</cmath> from which we see our equation becomes <cmath>\frac{2k+3}{k^2-1} = \frac{7}{51}, \ \ \implies \ \ 51(2k+3) = 7(k^2-1).</cmath>  
+
Note that the LHS equals <cmath>\bigg(\frac{2}{k} + \frac{2}{k^3} + \cdots \bigg) + \bigg(\frac{3}{k^2} + \frac{3}{k^4} + \cdots \bigg) = \frac{\frac{2}{k}}{1 - \frac{1}{k^2}} + \frac{\frac{3}{k^2}}{1 - \frac{1}{k^2}} = \frac{2k+3}{k^2-1},</cmath> from which we see our equation becomes <cmath>\frac{2k+3}{k^2-1} = \frac{7}{51}, \ \ \implies \ \ 51(2k+3) = 7(k^2-1).</cmath>  
  
 
Note that <math>17</math> therefore divides <math>k^2 - 1,</math> but as <math>17</math> is prime this therefore implies <cmath>k \equiv \pm 1 \pmod{17}.</cmath> (Warning: This would not be necessarily true if <math>17</math> were composite.) Note that <math>\boxed{\textbf{(D)} 16 }</math> is the only answer choice congruent satisfying this modular congruence, thus completing the problem. <math>\square</math>
 
Note that <math>17</math> therefore divides <math>k^2 - 1,</math> but as <math>17</math> is prime this therefore implies <cmath>k \equiv \pm 1 \pmod{17}.</cmath> (Warning: This would not be necessarily true if <math>17</math> were composite.) Note that <math>\boxed{\textbf{(D)} 16 }</math> is the only answer choice congruent satisfying this modular congruence, thus completing the problem. <math>\square</math>

Revision as of 22:43, 1 September 2023

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: $0.\overline{23}_k = 2\cdot k^{-1} + 3 \cdot k^{-2} + 2 \cdot k^{-3} + 3 \cdot k^{-4} + \cdots$

Notice that this is equivalent to \[2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + \cdots )\]

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

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 $\boxed{\textbf{(D) }16}$.

Solution 4

Assuming you are familiar with the rules for basic repeating decimals, $0.232323... = \frac{23}{99}$. Now we want our base, $k$, to conform to $23\equiv7\pmod k$ and $99\equiv51\pmod k$, the reason being that we wish to convert the number from base $10$ to base $k$. Given the first equation, we know that $k$ must equal 9, 16, 23, or generally, $7n+2$. The only number in this set that is one of the multiple choices is $16$. When we test this on the second equation, $99\equiv51\pmod k$, it comes to be true. Therefore, our answer is $\boxed{\textbf{(D) }16}$.

Solution 5

Note that the LHS equals \[\bigg(\frac{2}{k} + \frac{2}{k^3} + \cdots \bigg) + \bigg(\frac{3}{k^2} + \frac{3}{k^4} + \cdots \bigg) = \frac{\frac{2}{k}}{1 - \frac{1}{k^2}} + \frac{\frac{3}{k^2}}{1 - \frac{1}{k^2}} = \frac{2k+3}{k^2-1},\] from which we see our equation becomes \[\frac{2k+3}{k^2-1} = \frac{7}{51}, \ \ \implies \ \ 51(2k+3) = 7(k^2-1).\]

Note that $17$ therefore divides $k^2 - 1,$ but as $17$ is prime this therefore implies \[k \equiv \pm 1 \pmod{17}.\] (Warning: This would not be necessarily true if $17$ were composite.) Note that $\boxed{\textbf{(D)} 16 }$ is the only answer choice congruent satisfying this modular congruence, thus completing the problem. $\square$

~ Professor-Mom

Video Solution by OmegaLearn

https://youtu.be/SCGzEOOICr4?t=1110

~ pi_is_3.14

Video Solution 1

https://youtu.be/3YhYGSneu70

Education, the Study of Everything

Video Solution by WhyMath

https://youtu.be/W5V1UF-vSDE

~savannahsolver

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_18&oldid=197745"