One way to handle (1,2,3,4,5): 1, 2, 3, 4, 5 1, 3, 3, 4, 5 1, 3, 2, 4, 8 2, 4, 2, 4, 8 0, 4, 4, 4, 8 4, 4, 4, 4, 8 0, 8, 0, 8, 8 8, 8, 8, 8, 8
if we assume there is a solution and work backward, we will see that the result number must be in the form of 2^N provided that the 5 initial numbers are different. ie if we got x,x,x,x,x as the final line, By working backward, we have 0,x,x,x,x, x/2,x/2,x,x,x or 0,0,x,x,x ...
I don't know if there is a solution for any sets of 5 numbers. www.ddhw.com
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