Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2006 AIME I Problems/Problem 6

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 $\boxed{360}$

Solution 3

Another option is to do complementary counting and casework. $0.\overline{abc} = abc/999$. Excluding numbers with repeating digits our answer would just be $999(1000/2)/999 = 500$. We subtract off the sum of the numbers with at least two digits repeating:


Case 1: $aba$: Sum = $10(101 + 202 + 303 + \dots + 909) + 450(10) = 999(50)$.


Case 2: $aab$: Sum = $10(110 + 220 + 330 + \dots + 990) + 45(10) = 999(50)$.


Case 3: $baa$: Sum = $10(11 + 22 + 33 + \dots + 99) + 4500(10) = 999(50)$.


We overcounted the case where $a = b$ twice (the numbers with all 3 digits being the same). The sum of these numbers is $111 + 222 + 333 + \dots + 999 = 100(45) + 10(45) + 45 = 4995$.

So, our final answer is $(999 \cdot 500 - 999 \cdot 150 + 4995 \cdot 2)/999 = \boxed{360}$.

~sharmaguy

See also

2006 AIME I (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_6&oldid=251334"