AMC12 2014 A

AMC12 2014 A · Q23

AMC12 2014 A · Q23. It mainly tests Decimals, Digit properties (sum of digits, divisibility tests).

The fraction \[\dfrac1{99^2}=0.\overline{b_{n-1}b_{n-2}\ldots b_2b_1b_0},\] where $n$ is the length of the period of the repeating decimal expansion. What is the sum $b_0+b_1+\cdots+b_{n-1}$?

小数 \[\dfrac1{99^2}=0.\overline{b_{n-1}b_{n-2}\ldots b_2b_1b_0},\] 其中 $n$ 是循环小数展开的周期长度。求 $b_0+b_1+\cdots+b_{n-1}$ 的值？

(A) 874 874

(B) 883 883

(D) 891 891

(E) 892\qquad 892\qquad

Answer

Correct choice: (B)

正确答案：（B）

Solution

the fraction $\dfrac{1}{99}$ can be written as \[\sum^{\infty}_{n=1}\dfrac{1}{10^{2n}}\]. similarly the fraction $\dfrac{1}{99^2}$ can be written as $\sum^{\infty}_{m=1}\dfrac{1}{10^{2m}}\sum^{\infty}_{n=1}\dfrac{1}{10^{2n}}$ which is equivalent to \[\sum^{\infty}_{m=1}\sum^{\infty}_{n=1} \dfrac{1}{10^{2(m+n)}}\] and we can see that for each $n+m=k$ there are $k-1$ $(n,m)$ combinations so the above sum is equivalent to: \[\sum^{\infty}_{k=2}\dfrac{k-1}{10^{2k}}\] we note that the sequence starts repeating at $k = 102$ yet consider \[\sum^{101}_{k=99}\dfrac{k-1}{10^{2k}}=\dfrac{98}{{10^{198}}}+\dfrac{99}{{10^{200}}}+\dfrac{100}{10^{{202}}} \]\[=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{100}{10000})\]\[=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{1}{100})\]\[=\dfrac{1}{10^{198}}(98+\dfrac{100}{100})=\dfrac{1}{10^{198}}(99)\] so the decimal will go from 1 to 99 skipping the number 98 and we can easily compute the sum of the digits from 0 to 99 to be \[45\cdot10\cdot2=900\] subtracting the sum of the digits of 98 which is 17 we get \[900-17=883\textbf{(B) }\qquad\]

分数 $\dfrac{1}{99}$ 可写为 \[\sum^{\infty}_{n=1}\dfrac{1}{10^{2n}}\]。类似地，分数 $\dfrac{1}{99^2}$ 可写为 $\sum^{\infty}_{m=1}\dfrac{1}{10^{2m}}\sum^{\infty}_{n=1}\dfrac{1}{10^{2n}}$，等价于 \[\sum^{\infty}_{m=1}\sum^{\infty}_{n=1} \dfrac{1}{10^{2(m+n)}}\] 对于每个 $n+m=k$，有 $k-1$ 个 $(n,m)$ 组合，故上述和等价于： \[\sum^{\infty}_{k=2}\dfrac{k-1}{10^{2k}}\] 注意到序列在 $k = 102$ 处开始重复。然而考虑 \[\sum^{101}_{k=99}\dfrac{k-1}{10^{2k}}\]\[=\dfrac{98}{{10^{198}}}+\dfrac{99}{{10^{200}}}+\dfrac{100}{10^{{202}}}\]\[=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{100}{10000})\]\[=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{1}{100})\]\[=\dfrac{1}{10^{198}}(98+\dfrac{100}{100})\]\[=\dfrac{1}{10^{198}}(99)\] 故小数从 $1$ 到 $99$，跳过了数字 $98$。我们可以轻松计算 $0$ 到 $99$ 的各位数字和为 \[45\cdot10\cdot2=900\]，减去 $98$ 的各位数字和 $17$，得 \[900-17=883\textbf{(B) }\qquad\]

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