# 2014 AIME II Problems/Problem 4

Notice repeating decimals can be written as the following:

where a,b,c are the digits. Now we plug this back into the original fraction:

Multiply both sides by 999*99. This helps simplify the right side as well because 999=111*9=37*3*9:

Dividing both sides by 9 and simplifying gives:

At this point, seeing the 221 factor common to both a and b is crucial to simplify. This is because taking mod 221 to both sides results in:

Notice that we arrived to the result by simply dividing 9801 by 221 and seeing 9801=44*221+77. Okay, now it's pretty clear to divide both sides by 11 in the modular equation but we have to worry about 221 being multiple of 11. Well, 220 is a multiple of 11 so clearly, 221 couldn't be. Also, 221=13*17. Now finally we simplify and get:

But we know c is between 0 and 9 because it is a digit, so c must be 7. Now it is straightforward from here to find a and b:

and since a and b are both between 0 and 9, we have a=b=4. Finally we have the 3 digit integer