#3

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

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1 solution

Department 8
Sep 7, 2015

I used google for answering

1 x + 2 y = 4 2015 1 4 x + 1 2 y = 1 2015 \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 4 and other by 2 2 and using google I found there were 13 13 solutions to it out of which only 9 9 satisfied and a + b + c + d = 18 a+b+c+d = 18 .

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