2019 AMC 10A Problems/Problem 18

Revision as of 04:11, 18 January 2021 by Professor-mom (talk | contribs) (Solution 6)
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 legendary, amazingly, historically great calculations, this eventually gives us $\boxed{\textbf{(D) }16}$ as the answer.

-ellpet

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

Solution 5

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 6

Note that the LHS equals \[\bigg(\frac{2}{k} + \frac{2}{k^3} + \cdots \bigg) + \bigg(\frac{3}{k^2} + \frac{3}{k^4} + \cdots \bigg) = \frac{\frac{2}{k}}{1 - \frac{1}{k^2}} + \frac{\frac{3}{k^2}}{1 - \frac{1}{k^2}} = \frac{2k+3}{k^2},\] 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

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

Video Solution

https://youtu.be/3YhYGSneu70

Education, the Study of Everything

(Please put video solutions at the end in order of when they were edited in)

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

Invalid username
Login to AoPS