<s>[INST] 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
 [/INST] Your argument isn't correct as stated due to the issue pointed out by Dan Uznanski in the comments. But following your idea of combining things, we have
$$6abc = 7(ab + ac + bc)$$
so that $7 | abc$ and $6 | (ab + ac + bc)$. Let's assume that $a$ is divisible by $7$, so clearly $a \ge 7$ (assuming that all the numbers are positive). It then follows that
$$\frac 1 b + \frac 1 c = \frac 6 7 - \frac 1 a \ge \frac 5 7 > \frac 2 3$$
Now if $b$ and $c$ were both strictly greater than $2$, we would have a contradiction (why?). We can therefore assume $c = 2$ and reduce to
$$\frac 1 a + \frac 1 b = \frac 5 {14}.$$
Now repeat the ideas: We have
$$\frac 1 b = \frac 5 {14} - \frac 1 a \ge \frac 5 {14} - \frac 1 7 = \frac{3}{14} \implies b \le \frac{14}{3} \implies b \le 4.$$
This is now a very finite set to check.
</s>