# 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

• You can't deduce that one of the numbers must be $7$; all you can be sure of is that one of the numbers is a multiple of $7$. Having said that, I don't see a solution staring me in the face. – TonyK Apr 22 at 0:34
• @TonyK OMG you are right! I am having a brain glitch pfft... – Max0815 Apr 22 at 0:36

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. $$\frac12+\frac13+\frac1{42}=\frac67\\2+3+42=47$$

• Nice! But why must one of them be a multiple of $6$? (I mean, yes, there's only one solution, and one of them is a multiple of $6$; but how did you deduce it?) – TonyK Apr 22 at 1:00
• @John Doe I am wondering the same thing. – Max0815 Apr 22 at 1:01
• @TonyK yes, I think I got lucky there. I saw$$ab+ac+bc\equiv0\mod6$$But I realise now that that doesn't necessarily mean all the terms must be multiples of $6$... That being said, it may be possible to deduce by considering it case by case. E.g. if $ab\equiv 1,3,5\mod6$ then we arrive at a contradiction that the sum cannot be $\equiv0\mod6$. If instead $ab\equiv2\mod6$, then $ac$ and $bc$ also $\equiv2\mod6$ (same for $-2\equiv4$). I haven't bothered going further than that, but some more stuff can definitely be deduced (though I am not entirely sure if we get to the idea about multiples of 6) – John Doe Apr 22 at 1:29
• @JohnDoe very nice! I see! – Max0815 Apr 22 at 1:40
• @Max0815 To finish: Suppose $ab\equiv2\mod6$. We showed $bc,ac$ can't be $\equiv$ to an odd number, so its between $2,4\equiv-2$. The only combination of these leading to the total sum being $0$ is $2+2+2$. Then, $\implies a=2k_1,b=2k_2$. Then $2k_i c\equiv2\implies k_ic\equiv1,4\implies k_1\equiv k_2$. Also $4k_1k_2\equiv2\implies k_1k_2\equiv2,5$ but none of these numbers is a square number $\mod 6$ so this is not possible. The same argument holds for $ab\equiv4\equiv-2\mod6$. So we've shown $ab\not\equiv1,2,3,4,5\implies ab\equiv0\implies (b+a)c\equiv0\implies c\equiv0$ [Some steps omitted] – John Doe Apr 22 at 3:37

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.

• @TonyK Oh, I see. – Max0815 Apr 22 at 0:54