For some positive integer k, the repeating base- k representation of the (base-ten) fraction (7)/(51) is 0.\overline23_k = 0.232323\ldots_k. What is k?

2019 AMC 12A Problem 11 — Solution

2019 AMC 12A Problem 11

Attempt Problem 11 of the 2019 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 12A solutions, or check the answer key.

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11.

For some positive integer $k,$ the repeating base- $k$ representation of the (base-ten) fraction $\dfrac{7}{51}$ is $0.\overline{23}_k = 0.232323\ldots_k.$ What is $k?$

$13$

$14$

$15$

$16$

$17$

Answer: D

Concepts:number base repeating decimal quadratic

Difficulty rating: 1440

Solution:

The repeating block gives $0.\overline{23}_k = \dfrac{2k + 3}{k^2 - 1} = \dfrac{7}{51}.$

Cross-multiplying, $51(2k + 3) = 7(k^2 - 1),$ so $7k^2 - 102k - 160 = 0.$

The quadratic formula gives $\begin{aligned} k &= \dfrac{102 + \sqrt{14884}}{14} \\ &= \dfrac{102 + 122}{14} = 16. \end{aligned}$

Thus, the correct answer is D.

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