Three fractions with a numerator of 1 and denominators of $a, b$, and $c$ added together equals $\frac{6}{7}$. What is $a+b+c?$ 
If $a, b$ and $c$ are positive integers such that $\frac{1}{a} +
 \frac{1}{b} + \frac{1}{c} = \frac{6}{7}$ , then what is $a + b + c?$

I start by adding the LHS together, which results in $\frac{ab+bc+ca}{abc}=\frac{6x}{7x}$. I proceed with trial and error. I know the bottom is a multiple of 7, so thus one of the numbers must be 7. The top likewise has to be a multiple of 6. Taking me ~30 minutes to get to the 40th multiple of 7, doing this results in absolutely no progress. How would I solve this?
Thanks!
Max0815
 A: As you deduced, one of $a,b,c$ at least must be a multiple of $7$. Similarly, at least one of them must be a multiple of $6$, and the other two multiplied together must produce a multiple of $6$ (we don't end up having to use this information). Suppose $a=7k$. Then $\frac1a\le\frac17\implies \frac1b+\frac1c\ge\frac57\implies$ one of $b$ or $c$ must be $\le2$ (otherwise the sum of the reciprocals has no chance of reaching $\frac57$. But since the overall sum is $<1\implies a,b,c>1$. So wlog let $b=2$. Then, $$\frac1a+\frac1c=\frac5{14}$$
Then by the same argument, $\frac1c\ge\frac3{14}\implies c\le4\implies (c=3)\cup(c=4)$. Try these two, and we see $c=4$ does not work. $c=3$ gives $a=42$ as a solution. $$\frac12+\frac13+\frac1{42}=\frac67\\2+3+42=47$$
A: One of the numbers must be a multiple of $7$; say, $a=7k$ for some integer $k$. Then we get
$$\frac1b+\frac1c=\frac67-\frac{1}{7k}=\frac{6k-1}{7k}$$
To get rid of that $7$, we want $6k-1$ to be a multiple of $7$; so try $k=6$. This gives us
$$\frac1b+\frac1c=\frac{35}{42}=\frac56$$
This has an obvious solution.
The question remains whether it's the only solution.
