# 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.

• actual equation is $\frac{a*10+b}{99} + \frac{a*100+b*10+c}{999}=\frac{33}{37}$ Dec 10, 2016 at 20:21
• What are your thoughts on this problem? As a simpler problem, can you determine the fraction of $0.\overline{a}$? Dec 10, 2016 at 20:21

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.

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


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:

$$c = 7$$ $$ab = h = 11\times 4 = 44 \implies a=4,b=4$$