repeating decimals equation problem The repeating decimals $0.abab\overline{ab}$ and $.abcabc\overline{abc}$ satisfy
$0.abab\overline{ab} + 0.abcabc\overline{abc} = \frac{33}{37}$
where a,b, and c are (not necessarily distinct) digits. Find the three-digit number abc.
 A: HINT: Let $x=0.\overline{ab}$ and $y=0.\overline{abc}$. Then $100x=x+ab$, so $x=\frac{ab}{99}$, and, similarly, $y=\frac{abc}{999}$. You now know that
$$\frac{ab}{99}+\frac{abc}{999}=\frac{33}{37}\;.$$
Put everything over the common denominator of $10989$, and you find that
$$(111)(ab)+(11)(abc)=9801\;.$$
Now $(111)(ab)=ab00+ab0+ab$, and $(11)(abc)=abc0+abc$, so $a+a+\text{possible carry}=9$. Clearly there is a carry, and $a=4$. See if you can finish it from there.
An alternative (and definitely easier) approach is to use the same idea to notice that $x=\frac{ababab}{999999}$ and $y=\frac{abcabc}{999999}$, so that
$$ababab+abcabc=891891\;.$$
From here it’s easy to get $a=4$, and the other two letters then fall into places quite easily as well.
A: C code to solve this problem:
#include <stdio.h>

int main(void) {
    for (int a=0;a<10;a++) {
        for (int b=0;b<10;b++) {
            for (int c=0;c<10;c++) {
                int m = a*10+b;
                int n = a*100+b*10 + c;
                if (m/99.0+n/999.0==33/37.0) {
                    printf("a=%d b=%d c=%d\n",a,b,c);
                }
            }
        }
    }
    return 0;
}

Output:
a=4 b=4 c=7

A: The numbers can be expressed as $0.\overline {ab}$  and  $0.\overline {abc}$  and  written as fractions fractions as ${ 0.\overline {ab} }= {ab\over99} $ and $ 0.\overline {abc} = { abc\over999} $:
$$ {ab\over99} + { abc\over999} = {33\over 37} $$
Let $h = ab $ and $k = c \implies abc = 10h + k$
where $0\leq h<100  $ and $0 \leq k<10 $.
Rewriting the same expression using $h$ and $k$ :
 $$ {h\over99} + {10 h+k\over999} = {33\over 37} $$
We notice that $999 = 37 \times 3^3 $ and $99 = 3^2\times11$
$$ {h\over3^2\times11} + {10 h+k\over37 \times 3^3} = {33 \times 3^3 \over 37 \times 3^3 } $$
multiply by $3^2$:
$$ {h\over11} + {10 h+k\over37 \times 3} = {11 \times 3^3 \over 37  } $$
Since $h$ and $k$ are natural numbers, the only way this expression can be satisfied is by having $h$ to be divisible by 11 and $10h + k$ to be divisible by 3.
let $h = 11n$
where $n \in \{0,1,2,3,4,5,6,7,8,9\}  $ 
so 
$$ {n} + {10 \times 11 n +k\over37 \times 3} = {11 \times 3^3 \over 37  } $$
some algebra and we get
$$ 221 n + k = 891 $$
since $k<10$ and $221 \times 4 = 884$ we see that $n=4$ gets us close.
$$ 884 + k  = 891 \implies  k = 7$$
putting it all together:
$$c = 7$$
$$ab = h = 11\times 4 = 44 \implies a=4,b=4$$ 
