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

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==Problem==
 
==Problem==
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The fraction <cmath>\dfrac1{99^2}=0.\overline{b_{n-1}b_{n-2}\ldots b_2b_1b_0},</cmath> where <math>n</math> is the length of the period of the repeating decimal expansion.  What is the sum <math>b_0+b_1+\cdots+b_{n-1}</math>?
 
  
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<math>\textbf{(A) }874\qquad
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The fraction
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\textbf{(B) }883\qquad
 
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\textbf{(C) }887\qquad
 
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\textbf{(D) }891\qquad
 
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\textbf{(E) }892\qquad</math>
 
  
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== Solution ==
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<cmath>\dfrac1{99^2}=0.\overline{b_{n-1}b_{n-2}\ldots b_2b_1b_0},</cmath>
 +
 
 +
where <math>n</math> is the length of the period of the repeating decimal expansion.  What is the sum <math>b_0+b_1+\cdots+b_{n-1}</math>?
 +
 
 +
<math>\textbf{(A) }874\qquad \textbf{(B) }883\qquad \textbf{(C) }887\qquad \textbf{(D) }891\qquad \textbf{(E) }892\qquad</math>
 +
 
 +
==Solution 1==
  
 
the fraction <math>\dfrac{1}{99}</math> can be written as <cmath>\sum^{\infty}_{n=1}\dfrac{1}{10^{2n}}</cmath>.
 
the fraction <math>\dfrac{1}{99}</math> can be written as <cmath>\sum^{\infty}_{n=1}\dfrac{1}{10^{2n}}</cmath>.
Line 19: Line 20:
 
and we can easily compute the sum of the digits from 0 to 99 to be <cmath>45\cdot10\cdot2=900</cmath> subtracting the sum of the digits of 98 which is 17 we get
 
and we can easily compute the sum of the digits from 0 to 99 to be <cmath>45\cdot10\cdot2=900</cmath> subtracting the sum of the digits of 98 which is 17 we get
 
<cmath>900-17=883\textbf{(B) }\qquad</cmath>
 
<cmath>900-17=883\textbf{(B) }\qquad</cmath>
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==Solution 2==
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<math>\frac{1}{99^2}\\\\
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=\frac{1}{99} \cdot \frac{1}{99}\\\\
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=\frac{0.\overline{01}}{99}\\\\
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=0.\overline{00010203...9799}</math>
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 +
So, the answer is <math>0+0+0+1+0+2+0+3+...+9+7+9+9=2\cdot10\cdot\frac{9\cdot10}{2}-(9+8)</math> or <math>\boxed{\textbf{(B)}\ 883}</math>.
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There are two things to notice here. First, <math>\frac{1}{99}</math> has a very simple and unique decimal expansion, as shown. Second, for <math>\frac{0.\overline{01}}{99}</math> to itself produce a repeating decimal, <math>99</math> has to evenly divide a sufficiently extended number of the form <math>101010101..</math>. This number will have <math>99</math> ones (197 digits in total), as to be divisible by <math>9</math> and <math>11</math>. The enormity of this number forces us to look for a pattern, and so we divide out as shown. Indeed, upon division (seeing how the remainders always end in "501" or "601" or, at last, "9801"), we find the repeating part <math>.00010203040506070809101112131415....9799</math>. If we wanted to further check our pattern, we could count the total number of digits in our quotient (not counting the first three): 195. Since <math>99<100</math>, multiplying by it will produce either <math>1</math> or <math>2</math> extra digits, so our quotient passes the test.
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 +
Or note that the 1 has to be carried when you get to 100 so the 99 becomes 00 and the 1 gets carried again to 98 which becomes 99.
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== Video Solution by Richard Rusczyk ==
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 +
https://artofproblemsolving.com/videos/amc/2014amc12a/382
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2014|ab=A|num-b=22|num-a=24}}
 
{{AMC12 box|year=2014|ab=A|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 08:35, 26 May 2023

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 1

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 \[900-17=883\textbf{(B) }\qquad\]

Solution 2

$\frac{1}{99^2}\\\\ =\frac{1}{99} \cdot \frac{1}{99}\\\\ =\frac{0.\overline{01}}{99}\\\\ =0.\overline{00010203...9799}$

So, the answer is $0+0+0+1+0+2+0+3+...+9+7+9+9=2\cdot10\cdot\frac{9\cdot10}{2}-(9+8)$ or $\boxed{\textbf{(B)}\ 883}$.

There are two things to notice here. First, $\frac{1}{99}$ has a very simple and unique decimal expansion, as shown. Second, for $\frac{0.\overline{01}}{99}$ to itself produce a repeating decimal, $99$ has to evenly divide a sufficiently extended number of the form $101010101..$. This number will have $99$ ones (197 digits in total), as to be divisible by $9$ and $11$. The enormity of this number forces us to look for a pattern, and so we divide out as shown. Indeed, upon division (seeing how the remainders always end in "501" or "601" or, at last, "9801"), we find the repeating part $.00010203040506070809101112131415....9799$. If we wanted to further check our pattern, we could count the total number of digits in our quotient (not counting the first three): 195. Since $99<100$, multiplying by it will produce either $1$ or $2$ extra digits, so our quotient passes the test.

Or note that the 1 has to be carried when you get to 100 so the 99 becomes 00 and the 1 gets carried again to 98 which becomes 99.


Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2014amc12a/382

See Also

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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All AMC 12 Problems and Solutions

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