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Re: a problem (easy i think)

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  • rqrhn
    actually the second condition narrows the problem down. it means the number cant contain 4, 8, 9, 0 and also 6 if any of 2 or 3 is present. the numbers
    Message 1 of 2 , Dec 1, 2007
      actually the second condition narrows the problem down.

      it means the number cant contain 4, 8, 9, 0 and also 6 if any of 2
      or 3 is present.

      the numbers possible are 1, 2, 3, 5, 7 and 6 conditionally.

      so here are the solutions:

      get 2, 3 or 4 of the available digits.
      their product will always meet the criteon.
      then put 1 as many u wish so that the sum also matches the criterion.

      permutate it as u wish.


      123, 231, 312, 213, 321, 132
      11125, permutate it as u wish
      235

      127
      111156

      1167

      and so on,


      i see no bound
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