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 [[ | + | 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 == | + | == Solution 1 == |
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>. | 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>. | ||
+ | == Solution 2 == | ||
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}= \boxed{360}</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}= \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> | ||
== See also == | == See also == | ||
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[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:15, 9 February 2020
Contents
Problem
Let be the set of real numbers that can be represented as repeating decimals of the form where are distinct digits. Find the sum of the elements of
Solution 1
Numbers of the form can be written as . There are such numbers. Each digit will appear in each place value times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is .
Solution 2
Alternatively, for every number, , there will be exactly one other number, such that when they are added together, the sum is , or, more precisely, 1. As an example, .
Thus, the solution can be determined by dividing the total number of permutations by 2. The answer is .
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 . Then the total sum becomes which reduces to
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.