jmc

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 $999\ast 99.$ This helps simplify the right side as well because $999=111\ast 9=37\ast 3\ast 9$: $9990a+999b+9900a+990b+99c=33/37\ast 37\ast 3\ast 9\ast 99=33\ast 3\ast 9\ast 99$ 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 $mod221$ to both sides results in: $2210a+221b+11c\equiv 9801\mathrm{mod}221⟺11c\equiv 77\mathrm{mod}221$ Notice that we arrived to the result $9801\equiv 77\mathrm{mod}221$ by simply dividing $9801$ by $221$ and seeing $9801=44\ast 221+77.$ 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, $221=13\ast 17.$ Now finally we simplify and get: $c\equiv 7\mathrm{mod}221$ 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 $b$: $2210a+221b+11(7)=9801⟺221(10a+b)=9724⟺10a+b=44$ and since a and b are both between $0$ and $9$, we have $a=b=4$. Finally we have the $3$ digit integer $\boxed{447}$.