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Difference between revisions of "2014 AMC 12A Problems/Problem 23"

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== Solution ==
 
== Solution ==
  
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the fraction <math></math>\dfrac{1}{99}<math> can be written as </math>\sum^{\infty}_{n=1}\dfrac{1}{10^{-2n}}<math></math>.
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the fraction <math>\dfrac{1}{99}</math> can be written as <cmath>\sum^{\infty}_{n=1}\dfrac{1}{10^{-2n}}</cmath>.
 
similarly the fraction <math>\dfrac{1}{99^2}</math> can be written as <math>\sum^{\infty}_{m=1}\dfrac{1}{10^{-2m}}\sum^{\infty}_{n=1}\dfrac{1}{10^{-2n}}</math> which is equivalent to <cmath>\sum^{\infty}_{m=1}\sum^{\infty}_{n=1} \dfrac{1}{10^{-2(m+n)}}</cmath>
 
similarly the fraction <math>\dfrac{1}{99^2}</math> can be written as <math>\sum^{\infty}_{m=1}\dfrac{1}{10^{-2m}}\sum^{\infty}_{n=1}\dfrac{1}{10^{-2n}}</math> which is equivalent to <cmath>\sum^{\infty}_{m=1}\sum^{\infty}_{n=1} \dfrac{1}{10^{-2(m+n)}}</cmath>
 
and we can see that for each <math>n+m=k</math> there are <math>k-1</math>  <math>(n,m)</math> combinations so the above sum is equivalent to:
 
and we can see that for each <math>n+m=k</math> there are <math>k-1</math>  <math>(n,m)</math> combinations so the above sum is equivalent to:

Revision as of 21:19, 7 February 2014

Problem

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 $b_0+b_1+\cdots+b_{n-1}$?

$\textbf{(A) }874\qquad \textbf{(B) }883\qquad \textbf{(C) }887\qquad \textbf{(D) }891\qquad \textbf{(E) }892\qquad$

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 is repeating for at $k = 102$ yet consider \[\sum^{101}_{k=99}\dfrac{k-1}{10^{-2k}}=\dfrac{98}{10^{10^{198}}}+\dfrac{99}{10^{10^{200}}}+\dfrac{100}{10^{10^{202}}}=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{100}{10000})=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{1}{100})=\dfrac{1}{10^{198}}(98+\dfrac{100}{100})=\dfrac{1}{10^{198}}(99)\] 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 \[45*10*2=900\] subtracting the sum of the digits of 98 which is 17 we get \[900-17=883\textbf{(B) }\qquad\]

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