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Given the equation $$1/v + 1/w + 1/x + 1/y + 1/z = 1,$$ where all of $v,w,x,y,z$ are positive integers. One solution is the trivial $1/5 + 1/5 + 1/5 + 1/5 + 1/5$. For the minimum distinct solution for $v,w,x,y,z$. Find $v+w+x+y+z$.

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    $\begingroup$ What do you mean by the minimum distinct solution? Do you want to minimize $v+w+x+y+z$? $\endgroup$ – user17762 Apr 24 '13 at 16:53
  • $\begingroup$ As far as I know, 25 is the lowest value you can get, but I wouldn't know a formal proof, other than that if one DENOM is decreased by one, another DENOM has to be increased by more than 1, creating a sum more than 25. But admitting, this is no proof $\endgroup$ – imranfat Apr 24 '13 at 16:55
  • $\begingroup$ I think we are looking for the minimum of $x+y+z+u+v$ subject to $\frac1x+\frac1y+\frac1z+\frac1u+\frac1v=1$ and $(x,y,z,u,v)\ne(5,5,5,5,5)$. $\endgroup$ – Hagen von Eitzen Apr 24 '13 at 17:01
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    $\begingroup$ I think the problem wants all the denominators to be different, then to have minimum sum. A candidate is 2,4,6,18,36. $\endgroup$ – Ross Millikan Apr 24 '13 at 17:02
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    $\begingroup$ This is not a duplicate! $\endgroup$ – user17762 Apr 24 '13 at 17:03
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What is the minimum of $s:=x+y+z+u+v$ subject to $x<y<z<u<v$ and $\frac1x+\frac1y+\frac1z+\frac1u+\frac1v=1$?

From $\frac12+\frac14+\frac17+\frac1{14}+\frac1{28}=1$, we know thst $s\le 55$. Hence this is a finite problem. A quick brute force check of all possibilities with $x+y+z+u+v<55$ reveals that $$\tag{38} \frac13+\frac14+\frac15+\frac16+\frac1{20}=1$$ with $s=38$ is the optimum.

The next solutions with small sums are: $$\tag{43} \frac12+\frac14+\frac1{10}+\frac1{12}+\frac1{15}=1 $$ $$\tag{45} \frac12+\frac14+\frac1{9}+\frac1{12}+\frac1{18}= \frac12+\frac15+\frac1{6}+\frac1{12}+\frac1{20}=1 $$ $$\tag{50} \frac12+\frac14+\frac1{8}+\frac1{12}+\frac1{24}=1 $$

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If you want to find the minimum $v+w+x+y+z$ subject to $\dfrac1{v}+\dfrac1{w}+\dfrac1{x}+\dfrac1{y}+\dfrac1{z}=1$, here is one way.

From Cauchy-Schwarz, we have $$\left(\sqrt{v}\cdot\dfrac1{\sqrt{v}}+\sqrt{w}\cdot\dfrac1{\sqrt{w}}+\sqrt{x}\cdot\dfrac1{\sqrt{x}}+\sqrt{y}\cdot\dfrac1{\sqrt{y}}+\sqrt{z}\cdot\dfrac1{\sqrt{z}}\right)\leq\sqrt{v+w+x+y+z}\cdot\sqrt{\dfrac1v+\dfrac1w+\dfrac1x+\dfrac1y+\dfrac1z}$$ From this, conclude that $$v+w+x+y+z\geq 25$$ And you have the minimum hit for $(5,5,5,5,5)$.

If we want distinct numbers, we then have the following. (Clearly $25$ is a lower bound) $$(2,4,10,12,15)$$ which adds up to $43$.

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I suspect $2,4,8,12,24$ with sum $50$ will be hard to beat, but don't have a proof (yet).

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  • $\begingroup$ Answer is 38. I don't know how :'( $\endgroup$ – user74178 Apr 24 '13 at 17:17
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    $\begingroup$ $(2,4,10,12,15)$ beats it by $7$ :) $\endgroup$ – user17762 Apr 24 '13 at 17:22
  • $\begingroup$ Still answer is 38 ! $\endgroup$ – user74178 Apr 24 '13 at 17:32
  • $\begingroup$ @user74178: and it is shown in another answer. $\endgroup$ – Ross Millikan Apr 24 '13 at 17:36
  • $\begingroup$ What about $(3,4,5,6,20)$? It sums to $38$ and solves the equation. $\endgroup$ – Dejan Govc Apr 24 '13 at 17:36

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