Three numbers a, b, c, over 1 have a sum of $\frac{6}{7}$. What is a+b+c? Given that there are three integers $a, b,$ and $c$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{6}{7}$, what is the value of a+b+c?

Immediately, I see that I should combine the left hand side. Doing such results in the equation $$\frac{ab+ac+bc}{abc}=\frac{6x}{7x}.$$ This branches into two equation $$ab+ac+bc=6x$$$$abc=7x.$$ From this, I can tell that one of a, b, or c must be a multiple of 7, and the other two are a factor of $x$. Now, I do trial and error, but I find this very tiring and time consuming. Is there a better method?
Also, if you are nice, could you also help me on this($N$'s base-5 and base-6 representations, treated as base-10, yield sum $S$. For which $N$ are $S$'s rightmost two digits the same as $2N$'s?) question?
Thanks!
Max0815

Looks like the greedy algorithm wins here, and quite quickly.
Let's assume wlog that $a\le b\le c$.
If $a=1$, then we're already higher than $6/7$, so let's try $a=2$.  Now we have $\frac{1}{b} + \frac{1}{c} = \frac{6}{7} - \frac{1}{2} = \frac{5}{14}$.  Now, $b$ has to be at least $3$; let's try it, we get ... $\frac{1}{c} = \frac{5}{14} - \frac{1}{3} = \frac{1}{42}$.
Oh.
$$\frac{6}{7} = \frac{1}{2} + \frac{1}{3} + \frac{1}{42}$$
