Three numbers a, b, c, over 1 have a sum of $\frac{6}{7}$. What is a+b+c?

Question

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

You don't actually know that one of them must be 7. You know that one of them must be divisible by 7. — Dan Uznanski, Feb 14 '19 at 04:34

score 5 · Accepted Answer · answered Feb 14 '19 at 04:36

5

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.

answered Feb 14 '19 at 04:36

1

Was writing this up, but you beat me. Nice solution! – Jeremy Dover Feb 14 '19 at 04:37
Ok, I got it! Thanks! – Max0815 Feb 14 '19 at 04:42
Wait a sec I don't get why would we have contradiction if both b and c < 2? – Max0815 Feb 14 '19 at 04:46
@Max0815: It was a contradiction if $b$ and $c$ were both $gt 2$. The largest $\frac 1b+\frac1c$ could be would be $\frac 23$. – Ross Millikan Feb 14 '19 at 06:15

score 2 · Answer 2 · answered Feb 14 '19 at 04:39

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.

$$\frac{6}{7} = \frac{1}{2} + \frac{1}{3} + \frac{1}{42}$$

Three numbers a, b, c, over 1 have a sum of $\frac{6}{7}$. What is a+b+c?

2 Answers2