2019 AMC 10A Problem 18

Question

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..._k.

What is

k?

13

14

15

16

17

Expand

0.\overline{23}_k

as a geometric series

The common ratio after two digits is

k^{-2}

The repeating base-

k

fraction is

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

+2k^{-3}+3k^{-4}+\cdots

. Grouping odd and even powers gives

\begin{aligned} &2(k^{-1}+k^{-3}+\cdots) \\ &\quad {}+3(k^{-2}+k^{-4}+\cdots). \end{aligned}

Using geometric series, these sums are

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

and

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

, so

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

.

Setting

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

gives

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

, so

7k^2-102k-160=0

. Hence

k=16

. Thus, D is the correct answer.

Problem 18 in other years