From: petry@ix.netcom.com(david petry) Newsgroups: sci.math Subject: Re: Another number puzzle Date: 24 Mar 1996 21:51:05 GMT In asari@math.lfc.edu (ASARI Hirotsugu) writes: > >Can one produce 24 by using 3, 3, 7, 7? A friend of mine asked me and I >can't come up with one. Thanks. You didn't really say what the rules of your little game are, but how about 3*(7+7-3!) where 3! is the factorial of 3 (= 6) ============================================================================== From: mlerma@arthur.ma.utexas.edu (Miguel Lerma) Newsgroups: sci.math Subject: Re: Another number puzzle Date: 24 Mar 1996 23:02:44 GMT ASARI Hirotsugu (asari@math.lfc.edu) wrote: > Can one produce 24 by using 3, 3, 7, 7? A friend of mine asked me and I > can't come up with one. Thanks. 3! * (3 + 7/7) = 6 * 4 = 24 Miguel A. Lerma ============================================================================== From: David Shield Newsgroups: sci.math Subject: Re: Another number puzzle Date: 25 Mar 1996 09:10:08 GMT mlerma@arthur.ma.utexas.edu (Miguel Lerma) wrote: >ASARI Hirotsugu (asari@math.lfc.edu) wrote: >> Can one produce 24 by using 3, 3, 7, 7? A friend of mine asked me and I >> can't come up with one. Thanks. > > 3! * (3 + 7/7) = 6 * 4 = 24 > > >Miguel A. Lerma > also (3 * 7 / .7) - 3! = 30 - 6 = 24 David Shield ============================================================================== Newsgroups: sci.math From: rsherk@cln.etc.bc.ca (Ron Sherk) Subject: Re: Another number puzzle Date: Mon, 25 Mar 1996 03:53:55 GMT In a previous article, asari@math.lfc.edu (ASARI Hirotsugu) says: >Can one produce 24 by using 3, 3, 7, 7? A friend of mine asked me and I >can't come up with one. Thank you. A simple solution to this would be (7-3)!/(7-3!) = 4!/1 = 24. Hope this helps. Colin Sherk ============================================================================== From: hv@gold.compulink.co.uk (Hugo van der Sanden) Newsgroups: sci.math Subject: Re: Another number puzzle Date: 25 Mar 1996 11:11:56 GMT ASARI Hirotsugu (asari@math.lfc.edu) wrote: :Can one produce 24 by using 3, 3, 7, 7? A friend of mine asked me and I :can't come up with one. Thanks. What are the constraints? If the square root symbol can be used, you can have 3 + 3 * sqrt(7) * sqrt(7). Hugo van der Sanden ============================================================================== From: wfiguero@mosquito.com Newsgroups: sci.math Subject: Re: Another number puzzle Date: 25 Mar 1996 11:31:58 GMT since no restrictions were given try: 3!*3 + 7 - [ sqrt(7) ] where [ ] denotes the greatest integer function hence this leads to: 6*3 + 7 - 1 = 24 ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Another number puzzle Date: 26 Mar 1996 22:45:36 GMT In article , ASARI Hirotsugu wrote: >Can one produce 24 by using 3, 3, 7, 7? A friend of mine asked me and I >can't come up with one. Thanks. As others have pointed out this is sort of a poorly-defined question, but it has an interesting answer: 24 = 7 * ( 3 + 3/7 ) It's interesting because it uses only the four fundamental operations on integers and returns an integer answer, but requires a fractional intermediate result. (Thus when Wilbert Dijkhof (w.j.dijkhof@student.utwente.nl) posed this challenge once before I failed to find a solution using a program I had previously written.) dave