Consider the equation $\frac{1}{x} + \frac{2}{y} = \frac{4}{2015}$ for $x > 0 ; y > 0$ . Suppose the number of integer solution is A. Now $A^3 = 1000*a+100*b+10*c+d$ where $0<=a,b,c,d <=9$ . Find $a+b+c+d$ .

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18
11
17
13

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I used google for answering

$\frac { 1 }{ x } +\frac { 2 }{ y } =\frac { 4 }{ 2015 } \\ \frac { 1 }{ 4x } +\frac { 1 }{ 2y } =\frac { 1 }{ 2015 }$

This is an expression for Egyptian fractions , and one of terms to be divisible by $4$ and other by $2$ and using google I found there were $13$ solutions to it out of which only $9$ satisfied and $a+b+c+d = 18$ .