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)
(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.
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Posted by Daniel
on 2009-07-31 01:48:02 |
re(5): two views
