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How and When (Posted on 2009-07-29) Difficulty: 2 of 5
Solve this alphametic, where each of the capital letters in bold denotes a different decimal digit from 0 to 9. None of the numbers can contain any leading zero.

3√(HOW)+ 3√(AND) = 3√(WHEN)

No Solution Yet Submitted by K Sengupta    
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re(5): two views | Comment 12 of 17 |
(In reply to re(4): two views by brianjn)

well I think there is a simple solution to the precision problem.

looking at the equation

x^(1/3)+y^(1/3)=z^(1/3)  cube both sides

x+3(x^2y)^(1/3)+3(xy^2)^(1/3)+y=z

3(xy)^(1/3)*[x^(1/3)+y^(1/3)]=z-x-y brackets is qual to z^(1/3) so

3(xyz)^(1/3)=z-x-y

27xyz=(z-x-y)^3

so now we have an equivilant equation with no roots and if you plug in the values

x=192 y=375 and z=2187

both sides agree at the value 4251528000

thus I think this eliminates the possibility that this solution is a false positive also the secondary solution that was proposed fails to satisfy this equation thus can be shown not to be a solution.


  Posted by Daniel on 2009-07-31 01:48:02
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