On Tue, 8 Oct 2002, Ron Newman wrote: > On Tuesday, October 8, 2002, at 11:18 PM, Chris Devers wrote > > > 1 2 3 4 5 6 7 8 9 = 2002 > > > > The problem is to add any number of addition & multiplication > > operations wherever you'd like on the left such that in the > > end you have a valid equation. > > Do you have to keep the digits in the same order?
Yes. > If so, there are only 3^8=6561 solutions to try. Between each digit you > can have +, *, or nothing. Exactly. It can definitely be brute-forced, but I wouldn't be surprised if there was a cleverer way to get at the answer as well. I just haven't thought of it yet... -- Chris Devers [EMAIL PROTECTED] _______________________________________________ Boston-pm mailing list [EMAIL PROTECTED] http://mail.pm.org/mailman/listinfo/boston-pm
