AMC12 2004 A

AMC12 2004 A · Q25

AMC12 2004 A · Q25. It mainly tests Fractions, Base representation.

For each integer $n\geq 4$, let $a_n$ denote the base-$n$ number $0.\overline{133}_n$. The product $a_4a_5\cdots a_{99}$ can be expressed as $\frac {m}{n!}$, where $m$ and $n$ are positive integers and $n$ is as small as possible. What is $m$?

对每个整数 $n\geq 4$，令 $a_n$ 表示 $n$ 进制数 $0.\overline{133}_n$。乘积 $a_4a_5\cdots a_{99}$ 可表示为 $\frac {m}{n!}$，其中 $m$ 和 $n$ 为正整数，且 $n$ 尽可能小。求 $m$。

(A) 98 98

(B) 101 101

(D) 798 798

(E) 962 962

Answer

Correct choice: (E)

正确答案：（E）

Solution

This is an infinite geometric series with common ratio $\frac{1}{x^3}$ and initial term $x^{-1} + 3x^{-2} + 3x^{-3}$, so $a_x = \left(\frac{1}{x} + \frac{3}{x^2} + \frac{3}{x^3}\right)\left(\frac{1}{1-\frac{1}{x^3}}\right)$ $= \frac{x^2 + 3x + 3}{x^3} \cdot \frac{x^3}{x^3 - 1}$ $= \frac{x^2 + 3x + 3}{x^3 - 1}$ $= \frac{(x+1)^3 - 1}{x(x^3 - 1)}$. Alternatively, we could have used the algebraic manipulation for repeating decimals, \begin{align*} a_x &= \frac{1}{x}+\frac{3}{x^2}+\frac{3}{x^3}+\frac{1}{x^4}+\frac{3}{x^5}+\frac{3}{x^6}+\cdots \\ a_x \cdot x^3 &= x^2+3x+3+a_x\\ a_x(x^3-1) &= x^2+3x+3\\ a_x &= \frac{x^2+3x+3}{x^3-1}=\frac{(x+1)^3-1}{x(x^3-1)} \end{align*} Telescoping, \begin{align*} a_4a_5\cdots a_{99}&= \frac{(5^3-1)(6^3-1)\cdots (100^3-1)}{4 \cdot 5 \cdot 6 \cdot \cdots \cdot 99 \cdot (4^3-1)(5^3-1)\cdots(99^3-1)}\\ a_4a_5\cdots a_{99}&= \frac{999999}{4 \cdot 5 \cdot 6 \cdot \cdots \cdot 99 \cdot 63}=\frac{13 \cdot 37 \cdot 33 \cdot 6}{99!}\end{align*} Some factors cancel, (after all, $13 \cdot 37 \cdot 33 \cdot 6$ isn't one of the answer choices) \[\frac{13 \cdot 37 \cdot 33 \cdot 6}{99!}=\frac{13 \cdot 37 \cdot 2}{98!}\] Since the only factor in the numerator that goes into $98$ is $2$, $n$ is minimized. Therefore the answer is $13 \cdot 37 \cdot 2=962 \Rightarrow \text {(E)}$.

这是一个无穷等比级数，公比为 $\frac{1}{x^3}$，首项为 $x^{-1} + 3x^{-2} + 3x^{-3}$，所以 $a_x = \left(\frac{1}{x} + \frac{3}{x^2} + \frac{3}{x^3}\right)\left(\frac{1}{1-\frac{1}{x^3}}\right)$ $= \frac{x^2 + 3x + 3}{x^3} \cdot \frac{x^3}{x^3 - 1}$ $= \frac{x^2 + 3x + 3}{x^3 - 1}$ $= \frac{(x+1)^3 - 1}{x(x^3 - 1)}$。或者，我们也可以使用循环小数的代数处理方法， \begin{align*} a_x &= \frac{1}{x}+\frac{3}{x^2}+\frac{3}{x^3}+\frac{1}{x^4}+\frac{3}{x^5}+\frac{3}{x^6}+\cdots \\ a_x \cdot x^3 &= x^2+3x+3+a_x\\ a_x(x^3-1) &= x^2+3x+3\\ a_x &= \frac{x^2+3x+3}{x^3-1}=\frac{(x+1)^3-1}{x(x^3-1)} \end{align*} 利用望远镜相消， \begin{align*} a_4a_5\cdots a_{99}&= \frac{(5^3-1)(6^3-1)\cdots (100^3-1)}{4 \cdot 5 \cdot 6 \cdot \cdots \cdot 99 \cdot (4^3-1)(5^3-1)\cdots(99^3-1)}\\ a_4a_5\cdots a_{99}&= \frac{999999}{4 \cdot 5 \cdot 6 \cdot \cdots \cdot 99 \cdot 63}=\frac{13 \cdot 37 \cdot 33 \cdot 6}{99!}\end{align*} 有一些因子会约掉（毕竟，$13 \cdot 37 \cdot 33 \cdot 6$ 不在选项中） \[\frac{13 \cdot 37 \cdot 33 \cdot 6}{99!}=\frac{13 \cdot 37 \cdot 2}{98!}\] 由于分子中唯一能整除 $98$ 的因子是 $2$，因此 $n$ 被最小化。故答案为 $13 \cdot 37 \cdot 2=962 \Rightarrow \text {(E)}$。

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