# Difference between revisions of "2014 AIME II Problems/Problem 4"

(Created page with "Notice repeating decimals can be written as the following: <math>0.\overline{ab}=\frac{10a+b}{99}</math> <math>0.\overline{abc}=\frac{100a+10b+c}{999}</math> where a,b,c are t...") |
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+ | ==Problem== | ||

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+ | The repeating decimals <math>0.abab\overline{ab}</math> and <math>0.abcabc\overline{abc}</math> satisfy | ||

+ | |||

+ | <cmath>0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},</cmath> | ||

+ | |||

+ | where <math>a</math>, <math>b</math>, and <math>c</math> are (not necessarily distinct) digits. Find the three digit number <math>abc</math>. | ||

+ | |||

+ | ==Solution== | ||

Notice repeating decimals can be written as the following: | Notice repeating decimals can be written as the following: | ||

## Revision as of 15:15, 27 March 2014

## Problem

The repeating decimals and satisfy

where , , and are (not necessarily distinct) digits. Find the three digit number .

## Solution

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