2013 AMC 12A Problems/Problem 10
Problem
Let be the set of positive integers for which has the repeating decimal representation with and different digits. What is the sum of the elements of ?
Solution 1
Note that .
Dividing by 3 gives , and dividing by 9 gives .
The answer must be at least , but cannot be since no other than satisfies the conditions, so the answer is .
Solution 2
Let us begin by working with the condition . Let . So, . In order for this fraction to be in the form , must be a multiple of . Hence the possibilities of are . Checking each of these, and . So the only values of that have distinct and are and . So,
Solution 3
Notice that we have
We can subtract to get
From this we determine must be a positive factor of
The factors of are and .
For and however, they yield and which doesn't satisfy and being distinct.
For and we have and . (Notice that or can be zero)
The sum of these are
Solution 4
As in previous solutions, we have and . If we had , the decimal would be , which is characterized by and . So we seek the sum of the factors of 99 that are not also factors of 9.
Since , the sum is .
Video Solution
https://www.youtube.com/watch?v=XQpQaomC2tA
~sugar_rush
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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All AMC 12 Problems and Solutions |
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