AMC10 2019 A

AMC10 2019 A · Q18

AMC10 2019 A · Q18. It mainly tests Linear equations, Base representation.

For some positive integer $k$, the repeating base-$k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.23232323..._k$. What is $k$?

对于某个正整数 $k$，分数 $\frac{7}{51}$ 的基数-$k$ 循环表示为 $0.\overline{23}_k = 0.23232323..._k$。$k$ 是多少？

(A) 13 13

(B) 14 14

(D) 16 16

(E) 17 17

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The number $0.\overline{23}_k$ is the sum of an infinite geometric series with first term $\dfrac{2}{k}+\dfrac{3}{k^2}$ and common ratio $\dfrac{1}{k^2}$. Therefore the sum is $$ \frac{\dfrac{2}{k}+\dfrac{3}{k^2}}{1-\dfrac{1}{k^2}} =\frac{2k+3}{k^2-1} =\frac{7}{51}. $$ Then $0=7k^2-102k-160=(k-16)(7k+10)$, and therefore $k=16$.

答案（D）：数 $0.\overline{23}_k$ 是一个无穷等比级数的和，首项为 $\dfrac{2}{k}+\dfrac{3}{k^2}$，公比为 $\dfrac{1}{k^2}$。因此其和为 $$ \frac{\dfrac{2}{k}+\dfrac{3}{k^2}}{1-\dfrac{1}{k^2}} =\frac{2k+3}{k^2-1} =\frac{7}{51}. $$ 于是 $0=7k^2-102k-160=(k-16)(7k+10)$，因此 $k=16$。

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