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

(Deleted Solution #3 due to being exactly the same with Solution #2)
 
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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 <math>\frac{999}{2}</math>. Then the total sum becomes <math>\frac{\frac{999}{2}\times720}{999}</math> which reduces to <math>\boxed{360}</math>
 
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 <math>\frac{999}{2}</math>. Then the total sum becomes <math>\frac{\frac{999}{2}\times720}{999}</math> which reduces to <math>\boxed{360}</math>
 
== Solution 3 ==
 
By symmetry, the average over all <math>720=(10)(9)(8)</math> numbers is <math>\frac{1}{2}</math>. Then, their sum is <math>\frac{1}{2}720=\boxed{360}</math>.
 
  
 
== See also ==
 
== See also ==

Latest revision as of 18:14, 20 November 2023

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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