challenge for programmable calculators
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12-24-2013, 05:57 PM
Post: #33
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RE: challenge for programmable calculators
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 |
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