Difference between revisions of "2006 AIME I Problems/Problem 6"

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== Solution ==
 
== Solution ==
Numbers of the form <math>0.\overline{abc}</math> can be written as <math>\frac{abc}{999}</math>. There are <math>10\times9\times8=720</math> such numbers. Each digit will appear in each place value <math>\frac{720}{10}=72</math> times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is <math>\frac{45\times72\times111}{999}=360</math>.
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Numbers of the form <math>0.\overline{abc}</math> can be written as <math>\frac{abc}{999}</math>. There are <math>10\times9\times8=720</math> such numbers. Each digit will appear in each place value <math>\frac{720}{10}=72</math> times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is <math>\frac{45\times72\times111}{999}= \boxed{360} </math>.
  
 
Alternatively, for every number, <math>0.\overline{abc}</math>, there will be exactly one other number, such that when they are added together, the sum is <math>0.\overline{999}</math>, or, more precisely, 1. As an example, <math>.\overline{123}+.\overline{876}=.\overline{999} \Longrightarrow 1</math>.
 
Alternatively, for every number, <math>0.\overline{abc}</math>, there will be exactly one other number, such that when they are added together, the sum is <math>0.\overline{999}</math>, or, more precisely, 1. As an example, <math>.\overline{123}+.\overline{876}=.\overline{999} \Longrightarrow 1</math>.
  
Thus, the solution can be determined by dividing the total number of [[permutation]]s by 2. The answer is <math>\frac{10 \cdot 9 \cdot 8}{2} = \frac{720}{2}= 360</math>.
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Thus, the solution can be determined by dividing the total number of [[permutation]]s by 2. The answer is <math>\frac{10 \cdot 9 \cdot 8}{2} = \frac{720}{2}= \boxed{360}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 09:02, 31 August 2012

Problem

Let $\mathcal{S}$ be the set of real numbers that can be represented as repeating decimals of the form $0.\overline{abc}$ where $a, b, c$ are distinct digits. Find the sum of the elements of $\mathcal{S}.$

Solution

Numbers of the form $0.\overline{abc}$ can be written as $\frac{abc}{999}$. There are $10\times9\times8=720$ such numbers. Each digit will appear in each place value $\frac{720}{10}=72$ times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is $\frac{45\times72\times111}{999}= \boxed{360}$.

Alternatively, for every number, $0.\overline{abc}$, there will be exactly one other number, such that when they are added together, the sum is $0.\overline{999}$, or, more precisely, 1. As an example, $.\overline{123}+.\overline{876}=.\overline{999} \Longrightarrow 1$.

Thus, the solution can be determined by dividing the total number of permutations by 2. The answer is $\frac{10 \cdot 9 \cdot 8}{2} = \frac{720}{2}= \boxed{360}$.

See also

2006 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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