A Fisherman's Boxes
Level
pending
A fisherman had 2 unequal cubical boxes with rational side-lengths whose contents total exactly 6 cubic metres. Writing the side-length of boxes 1 and 2 as a/b and c/d respectively, where gcd(a,b)=1 and gcd(c,d)=1, what is a+b+c+d?
The answer is 96.
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1 solution
We have to solve (a/b)^3 + (c/d)^3 = 6 for positive integral values of a, b, c, d.
Multiplying the above equation by b^3, we get a^3 + {(c^3) (b^3)/(d^3)} = 6 (b^3). The RHS is an integer and to make it possible, d^3 must divide (c^3)*(b^3) i.e. d^3 must divide b^3 because gcd(c,d) = 1. Therefore, d must divide b.
Multiplying the above equation by d^3, we get {(a^3) (d^3)/(b^3)} + c^3 = 6 (d^3). If we proceed similarly as in the previous case, we get that b^3 must divide d^3 ==> b must divide d.
Thus, we get b = d and so, the equation transforms into: a^3 + c^3 = 6*(b^3). I'm yet to solve this equation mathematically but using programming, I get a = 37, b = 17, c = 21 = d.
a + b + c + d = 37 + 21 + 17 + 21 = 96.
Eagerly waiting for a better answer.