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 | + | 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 == | ||
+ | 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>. | ||
+ | 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 == |
− | + | {{AIME box|year=2006|n=I|num-b=5|num-a=7}} | |
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− | + | [[Category:Intermediate Combinatorics Problems]] | |
− | + | [[Category:Intermediate Number Theory Problems]] | |
+ | {{MAA Notice}} |
Latest revision as of 18:14, 20 November 2023
Contents
[hide]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 |
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