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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

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  • $\begingroup$ 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. $\endgroup$ – TonyK Apr 22 at 0:34
  • $\begingroup$ @TonyK OMG you are right! I am having a brain glitch pfft... $\endgroup$ – Max0815 Apr 22 at 0:36
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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. $$\frac12+\frac13+\frac1{42}=\frac67\\2+3+42=47$$

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    $\begingroup$ 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?) $\endgroup$ – TonyK Apr 22 at 1:00
  • $\begingroup$ @John Doe I am wondering the same thing. $\endgroup$ – Max0815 Apr 22 at 1:01
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    $\begingroup$ @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) $\endgroup$ – John Doe Apr 22 at 1:29
  • $\begingroup$ @JohnDoe very nice! I see! $\endgroup$ – Max0815 Apr 22 at 1:40
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    $\begingroup$ @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] $\endgroup$ – John Doe Apr 22 at 3:37
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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.

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

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