1992 AIME Problems/Problem 5
Problem
Write 
without using decimals
Solution
We consider the method in which repeating decimals are normally converted to fractions with an example:
Thus, let 
If 
is not divisible by 
or 
, then this is in lowest terms. Let us consider the other multiples: 
multiples of , 
of , and 
of and , so 
, which is the amount that are neither. The 
numbers that are multiples of 
reduce to multiples of . We have to count these since it will reduce to a multiple of which we have removed from 
, but, this cannot be removed since the numerator cannot cancel the .There aren't any numbers which are multiples of 
, so we can't get numerators which are multiples of . Therefore 
.
Solution 2
We can see that the denominator of the fraction will be 999b, where b is either 1, 1/3, 1/111, 1/9, or 1/333. This means that the numerator stays abc unless abc divides 999. Finding the numbers relatively prime to 999 gives us the numerators when the fraction cannot be simplified past 999.
Now, look at a number that does divide 999:
We can see that if a number does divide 999, it will be simplified into a number which is relatively prime to 999 (since that is what simplifying is). There are 660 numbers less than or relatively prime to 999, so the answer is 
.
- krishkhushi09
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