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

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== Problem ==
 
== Problem ==
Let <math> \mathcal{S} </math> be the set of [[real number]]s that can be represented as repeating [[decimal notation| decimals]] of the form <math> 0.\overline{abc} </math> where <math> a, b, c </math> are distinct [[digit]]s. Find the sum of the elements of <math> \mathcal{S}. </math>
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Let <math> \mathcal{S} </math> be the set of [[real number]]s that can be represented as repeating [[Decimal| decimals]] of the form <math> 0.\overline{abc} </math> where <math> a, b, c </math> are distinct [[digit]]s. Find the sum of the elements of <math> \mathcal{S}. </math>
  
 
== Solution 1 ==
 
== Solution 1 ==

Revision as of 00:15, 10 February 2020

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 1

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 $\frac{45\times72\times111}{999}= \boxed{360}$.

Solution 2

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 $\frac{10 \cdot 9 \cdot 8}{2} = \frac{720}{2}= \boxed{360}$.

Another method, albeit a little risky, that can be used is to note that the numbers between 1 and 999 with distinct digits average out to $\frac{999}{2}$. Then the total sum becomes $\frac{\frac{999}{2}\times720}{999}$ which reduces to $\boxed{360}$

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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