2019 AMC 10A Problems/Problem 18

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

Solution 2

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

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

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 $\boxed{D}$ .

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

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.

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