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 == | + | == 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>. | ||
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| + | == 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>. | ||
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| + | 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>. | ||
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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> | ||
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| + | ==Solution 3== | ||
| + | Another way is to do complementary counting and casework. <math>0.\overline{abc} = abc/999</math>. Excluding numbers with repeating digits our answer would just be <math>\frac{999 \cdot (1000/2)}{999} = 500</math>. We subtract off the sum of the numbers with at least two digits being the same: | ||
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| + | Case 1: <math>aba</math>: Sum = <math>1010(1 + 2 + 3 + \dots + 9) + 450\cdot 10 = 999 \cdot 50</math>. | ||
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| + | Case 2: <math>aab</math>: Sum = <math>1100(1 + 2 + 3 + \dots + 9) + 45\cdot 10 = 999 \cdot 50</math> | ||
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| + | Case 3: <math>baa</math>: Sum = <math>110(1 + 2 + 3 + \dots + 9) + 4500\cdot 10 = 999 \cdot 50</math>. | ||
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| + | We overcounted the case where <math>a = b</math> '''twice''' (the numbers with all 3 digits being the same). The sum of these numbers is <math>111(1 + 2 + 3 + \dots + 9) = 999 \cdot 5</math>. | ||
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| + | So, our final answer is <math>\frac{999 \cdot 500 - 999 \cdot 150 + 999 \cdot 5 \cdot 2}{999} = \boxed{360}</math>. | ||
| + | |||
| + | ~[https://artofproblemsolving.com/wiki/index.php/User:grogg007 grogg007] | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=2006|n=I|num-b=5|num-a=7}} | |
| + | |||
| + | [[Category:Intermediate Combinatorics Problems]] | ||
| + | [[Category:Intermediate Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 01:50, 31 July 2025
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 
Solution 3
Another way is to do complementary counting and casework. 
. Excluding numbers with repeating digits our answer would just be 
. We subtract off the sum of the numbers with at least two digits being the same:
Case 1: 
: Sum = 
.
Case 2: 
: Sum = 
Case 3: 
: Sum = 
.
We overcounted the case where 
twice (the numbers with all 3 digits being the same). The sum of these numbers is 
.
So, our final answer is 
.
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. 
