challenge for programmable calculators

[radwilliams]

I think I've found an equivalent numerical solution but not an identical solution (the order of operations is critical otherwise it is 0=0 and that is not a valid solution).

If you first expand the left side of (a+b+c)*(a*b*c) = abc

a^2+b^2+c^2 + 2*(a*b + a*c + b*c) = abc, now let a=b=0 and c=1

0^2 + 0^2 + 1^2 + 2*( 0 + 0 + 0 ) = 001

1 = 001 could be a solution it all depends on if '1' alone is the same as 001? Food for thought.

Happy Holidays to all, Ronald Williams