2014 AIME II Problems/Problem 4

Notice repeating decimals can be written as the following:

$0.\overline{ab}=\frac{10a+b}{99}$

$0.\overline{abc}=\frac{100a+10b+c}{999}$

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

$\frac{10a+b}{99}+\frac{100a+10b+c}{999}=\frac{33}{37}$

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

$9990a+999b+9900a+990b+99c=33/37*37*3*9*99=33*3*9*99$

Dividing both sides by 9 and simplifying gives:

$2210a+221b+11c=99^2=9801$

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:

$2210a+221b+11c \equiv 9801 \mod 221 \iff 11c \equiv 77 \mod 221$

Notice that we arrived to the result $9801 \equiv 77 \mod 221$ 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:

$c \equiv 7 \mod 221$

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:

$2210a+221b+11(7)=9801 \iff 221(10a+b)=9724 \iff 10a+b=44$

and since a and b are both between 0 and 9, we have a=b=4. Finally we have the 3 digit integer $\boxed{447}$

Page

Toolbox

Search

2014 AIME II Problems/Problem 4