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}

0.\overline{23}_k = 0.232323\ldots_k.

What is

k?

13

14

15

16

17

A two-digit repeating block in base

k

equals

\dfrac{2k + 3}{k^2 - 1}

Set

\dfrac{2k + 3}{k^2 - 1} = \dfrac{7}{51}

and clear denominators

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),

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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