2019 AMC 12A 第 11 题

Question

对某个正整数

k

，分数

\dfrac{7}{51}

的

k

进制循环表示为

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

。求

k

？

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

k

进制中两位循环节的值为

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

。

A two-digit repeating block in base

k

equals

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

令

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

，然后清分母

Set

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

and clear denominators

循环节给出

0.\overline{23}_k = \dfrac{2k + 3}{k^2 - 1} = \dfrac{7}{51}\text{。}

交叉相乘得

51(2k + 3) = 7(k^2 - 1)

，所以

7k^2 - 102k - 160 = 0

。

由二次公式，

\begin{aligned} k &= \dfrac{102 + \sqrt{14884}}{14} \\ &= \dfrac{102 + 122}{14} = 16 \end{aligned}\text{。}

所以正确答案是 D。

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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