The following solution assumes that brackets may be used with the given operators:
Where F(n1) > 1, F(n1) added to F(n1)/F(n1) iterated [F(n)  F(n1)] times should produce the value F(n).
Remaining will be either an odd or even number of F(n1)'s. If even, the number of F(n1)'s remaining can be paired as either zeroes {[F(n1)  F(n1)]} to be added to the total, or ones
{F(n1)/F(n1)} to be multiplied to the total. If odd, the quanity of F(n1)'s remaining will be >=5 [except for F(n1)=5]. In these odd cases, one can multiply the odd number of the "fibs"
{[F(n1)+F(n1)]/F(n1)  [F(n1)/F(n1)]} to the total. For the special case where F(n1) = 5 {F(n) = 8}, i.e., 8 5's, there is the solution 8 = (5*5+5)/5 + 5/5 + 5/5.

Posted by Dej Mar
on 20100710 10:42:21 