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

(Solution)
(Solution)
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Alternatively, for every number, .abc there will be exactly one other number, such that when it is summed to .abc, it will equal .999, or, more precisely, 1.
 
Alternatively, for every number, .abc there will be exactly one other number, such that when it is summed to .abc, it will equal .999, or, more precisely, 1.
  
Ex. <math>.123+.876=.999 -> 1 </math>
+
Ex. <math>.123+.876=.999   -> 1 </math>
  
 
Thus, the solution can be determined by dividing the total number of permutations by 2.
 
Thus, the solution can be determined by dividing the total number of permutations by 2.

Revision as of 11:47, 13 March 2007

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

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 $\frac{45\times72\times111}{999}=360$.

Alternatively, for every number, .abc there will be exactly one other number, such that when it is summed to .abc, it will equal .999, or, more precisely, 1.

Ex. $.123+.876=.999   -> 1$

Thus, the solution can be determined by dividing the total number of permutations by 2.

$\frac{720}{2}=360$

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