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So this is how I approached this question, the above equations could be simplified to :

$$a = \frac{4(b+c)}{b+c+4}\tag{1!}$$

$$b = \frac{10(a+c)}{a+c+10}\tag{2}$$

$$c=\frac{56(a+b)}{a+b+56}\tag{3}$$ From above, we can deduce that $4 > a$ since $\frac{(b+c)}{b+c+4} < 1$ similarly $10 > b, 56 > c$ so $a + b + c < 70$

Let, $$(a + b + c)k = 70\tag4$$

Now let, $$\alpha(b+c) = b+c+4\tag{1'}$$

$$\beta(a+c) = a+c+10\tag{2'}$$

$$\gamma( a+b ) = a+b+56\tag{3'}$$

Now adding the above 3 equations we get :

$$2(a+b+c) + 70 = a(\gamma + \beta) + c(\beta + \alpha) + b(\alpha + \gamma) \rightarrow (2 + k)(a+b+c) = a(\gamma + \beta) + c(\beta + \alpha) + b(\alpha + \gamma)$$

Now from above we see that coefficient of $a,b,c$ must be equal on both sides so, $$(2 + k) = (\alpha + \beta) = (\beta + \gamma) = (\alpha + \gamma)$$

Which implies $\beta = \gamma = \alpha = 1+ \frac{k}{2} = \frac{2 + k}{2}$, Now from $(1)$ and $(1')$ we get $a = \frac{4}{\alpha} = \frac{8}{2+k}$ similarly from $(2),(2')$ and $(3),(3')$ we find, $b = \frac{20}{2+k}, c = \frac{112}{2+k}$

Thus from above we get $a+b+c = \frac{140}{2+k}$ and from $(4)$ we get: $\frac{140}{2+k} = \frac{70}{k}$ from which we can derive $k = 2$

Thus we could derive $a = 2, b = 5, c = 28$ but, the problem now is $a, b, c$ values don't satisfy equation $(4)$ above for $k =2$

Well so, where do I err ? And did I take the right approach ? Do post the solution about how you solved for $x$.

  • $\begingroup$ @Watson it isn't like that, look I have asked atleast many such questions here and got answers, then I didn't had the scholar badge(which permits me to accept any kind of given answer), so then I showed my thanksgiving by upvoting those answers, now to revisit all of my previous questions and 'accepting' those answers would be a tedious task. $\endgroup$ – Arnav Das Jun 19 '16 at 9:45
  • $\begingroup$ No problem! I can help you to find answers you could maybe accept : (1), (2), (3), (4), … $\endgroup$ – Watson Jun 19 '16 at 9:50
  • $\begingroup$ (My aim is not to criticize you ! This is just to get the answers accepted :-) ). $\endgroup$ – Watson Jun 19 '16 at 9:51
  • $\begingroup$ Oh no by no way I took offense, just a little lazy I am.....but I guess now I don't have to $\endgroup$ – Arnav Das Jun 19 '16 at 10:27
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    $\begingroup$ Thank you! I fully understand that it can be tedious. But it would be great if you could accept answers you find useful. This was just to tell you. Sorry for being a bit annoying ;-). $\endgroup$ – Watson Jun 19 '16 at 11:24

Your deductions are wrong and that is what is misleading you. Integers can be both positive and negative.

If you solve equations (1), (2) and (3) simultaneously you can find a, b and c. I did this to find $$a=3$$ $$b=5$$ $$c=7$$

You can then plug this into the forth equation given in the problem to solve for x. $$x = \frac{abc}{ a + b + c} $$

which solves to give $$x=\frac{105}{15}=7$$

  • $\begingroup$ Well by 'simultaneously' do you mean, like putting values $a$ and $b$ of equation 1, 2 directly into equation 3 ? $\endgroup$ – Arnav Das Apr 26 '16 at 10:29
  • $\begingroup$ Your expression for $x$, Soph, comes down to $abc/(a+b+c)$. So, if it's true that $a=3$, $b=5$, and $c=7$, then the value you get for $x$ is quite wrong. $\endgroup$ – Gerry Myerson Apr 26 '16 at 11:08
  • $\begingroup$ Exactly, this is very long so I just used wolfram alfa to solve. But since there are 3 equations with 3 unknowns it is solvable. It would be quicker as Joshua suggested to find a trick to find values for 'a+b+c' and 'abc' but this is the more brute force solution... $\endgroup$ – Soph Apr 26 '16 at 11:09
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    $\begingroup$ Whoa I made a huge typo thanks @GerryMyerson now solution is much simpler. thanks! $\endgroup$ – Soph Apr 26 '16 at 11:19
  • $\begingroup$ Well I thought about going your way, but I evaded it for I thought it would be too tedious, ok if you were to modify my method, what would you do in it ? $\endgroup$ – Arnav Das Apr 26 '16 at 12:49

You seem to be assuming through out that $a$,$b$,$c$ are positive integers (e.g. in your deduction that $a < 4$), which doesn't appear to be in the problem statement.

Given they're all integers, I would clear denominators so you don't have to deal with fractions. We then notice that the last equation becomes $x(a+b+c) = abc$, so you don't need to determine the actual values of $a,b,c$ but rather their sum and product.

I haven't checked what happens when you clear denominators in the top three equations but I would hope that by taking linear combinations of them you can obtain an expression for $a+b+c$ and $abc$.

  • $\begingroup$ Hmm.....you are right.....well can you suggest another way to approach this problem $\endgroup$ – Arnav Das Apr 26 '16 at 10:12
  • $\begingroup$ I have updated my answer with a suggestion. $\endgroup$ – Josh Hunt Apr 26 '16 at 10:20
  • $\begingroup$ Ohk....lemme try this way.... $\endgroup$ – Arnav Das Apr 26 '16 at 10:22

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