Difference between revisions of "2006 AIME I Problems/Problem 6"
(→Solution) |
(→Solution) |
||
Line 10: | Line 10: | ||
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}=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}=360</math>. | ||
− | Alternatively, for every number, .abc there will be exactly one other number, such that when | + | Alternatively, for every number, .abc there will be exactly one other number, such that when they are added together, the sum will equal <math>0.\overline{.999}</math>, or, more precisely, 1. |
Ex. <math>.123+.876=.999 </math> -> 1 | Ex. <math>.123+.876=.999 </math> -> 1 |
Revision as of 10:48, 13 March 2007
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
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 .
Alternatively, for every number, .abc there will be exactly one other number, such that when they are added together, the sum will equal , or, more precisely, 1.
Ex. -> 1
Thus, the solution can be determined by dividing the total number of permutations by 2.
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 |