AMC12 2016 B

AMC12 2016 B · Q22

AMC12 2016 B · Q22. It mainly tests Primes & prime factorization, Remainders & modular arithmetic.

For a certain positive integer $n$ less than $1000$, the decimal equivalent of $\frac{1}{n}$ is $0.\overline{abcdef}$, a repeating decimal of period $6$, and the decimal equivalent of $\frac{1}{n+6}$ is $0.\overline{wxyz}$, a repeating decimal of period $4$. In which interval does $n$ lie?

对于某个小于 $1000$ 的正整数 $n$，$\frac{1}{n}$ 的小数表示为 $0.\overline{abcdef}$，这是一个循环节为 $6$ 的循环小数；而 $\frac{1}{n+6}$ 的小数表示为 $0.\overline{wxyz}$，这是一个循环节为 $4$ 的循环小数。问：$n$ 位于哪个区间内？

(A) [1, 200] [1, 200]

(B) [201, 400] [201, 400]

(D) [601, 800] [601, 800]

(E) [801, 999] [801, 999]

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Because $\frac{1}{n}=\frac{abcdef}{999999}$, it follows that $n$ is a divisor of $10^6-1=(10^3-1)(10^3+1)=3^3\cdot 7\cdot 11\cdot 13\cdot 37$. Because $\frac{1}{n+6}=\frac{wxyz}{9999}$, it follows that $n+6$ divides $10^4-1=3^2\cdot 11\cdot 101$. However, $n+6$ does not divide $10^2-1=3^2\cdot 11$, because otherwise the decimal representation of $\frac{1}{n+6}$ would have period 1 or 2. Thus $n=101k-6$, where $k=1,3,9,11,33,$ or $99$. Because $n<1000$, the only possible values of $k$ are 1, 3, and 9, and the corresponding values of $n$ are 95, 297, and 903. Of these, only $297=3^3\cdot 11$ divides $10^6-1$. Thus $n\in[201,400]$. It may be checked that $\frac{1}{297}=0.\overline{003367}$ and $\frac{1}{303}=0.\overline{0033}$.

答案（B）：因为$\frac{1}{n}=\frac{abcdef}{999999}$，可知$n$是$10^6-1=(10^3-1)(10^3+1)=3^3\cdot 7\cdot 11\cdot 13\cdot 37$的因数。因为$\frac{1}{n+6}=\frac{wxyz}{9999}$，可知$n+6$整除$10^4-1=3^2\cdot 11\cdot 101$。然而，$n+6$不整除$10^2-1=3^2\cdot 11$，否则$\frac{1}{n+6}$的小数表示将以1或2为循环节。因此$n=101k-6$，其中$k=1,3,9,11,33$或$99$。因为$n<1000$，$k$唯一可能的取值为1、3和9，相应的$n$为95、297和903。在这些数中，只有$297=3^3\cdot 11$整除$10^6-1$。因此$n\in[201,400]$。可检验$\frac{1}{297}=0.\overline{003367}$且$\frac{1}{303}=0.\overline{0033}$。

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