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Re: Problem 320

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  • cooperpuzzles
    ... 2.7027027.... being very close to e. ... 2^11*3^26=5205741216417792 2^8*3^28 =5856458868470016 2^5*3^30 =6588516227028768 2^2*3^32 =7412080755407364 Each
    Message 1 of 15 , Mar 1 1:23 AM
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      --- In mathforfun@yahoogroups.com, clooneman@y... wrote:
      >

      > But if the addends could contain a decimal part, then they would
      >all
      > equal n/d, and you'd have:
      > 2.7027027....*2.7027027....*2.7027027....*2.7027027.... (37 times)
      > = 2.7027027....^37 = 9 474 061 716 781 832.651871822612603, as
      >above.
      2.7027027.... being very close to e.

      >
      > P.S. Only one problem with my answer if n/d is an integer, because
      > using 36 integers (three less 2s and two more 3's) yields 5 856 458
      > 868 470 016. Whoops....
      >
      > P.P.S. And repeating that step again gets you a bigger product
      > again....
      2^11*3^26=5205741216417792
      2^8*3^28 =5856458868470016
      2^5*3^30 =6588516227028768
      2^2*3^32 =7412080755407364

      Each new line is the previous divided by 8 and multiplied by 9.

      >But interestingly, two 2's and thirty-two 3's gets you
      > exactly the same result as when you shrink the number of integers
      >by
      > 1 and use thirty-two 3's and one 4! Anyone see why?

      duh. Because 2+2=4?

      This is interesting. When I did this problem before, I always used
      rationals for addends. Rational addends close to e always gave the
      best results. So when Peter used natural numbers for addends, I
      thought they must be 3's and 2's because they are closest to e. Now
      you are showing that using a 4 can give the same result. Four is not
      very close to e.

      btw
      e^(100/e)=9479842689868732.4631217804012218
      100/e=36.7879441171442321595523770161461

      I can add e 36.7879.. times.
      Can I multiply e 36.7879.. times?

      >
      (Phew......)^2
      >
    • Peter Otzen
      ... More importantly 2x2=4 ... not very close to e. The addends must be 2 or 3, not because they are closer to e, but because an alternative to the addend 5,
      Message 2 of 15 , Mar 1 3:06 AM
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        --- In mathforfun@yahoogroups.com, "cooperpuzzles"
        <cooperpuzzles@y...> wrote:
        >
        > --- In mathforfun@yahoogroups.com, clooneman@y... wrote:
        > >
        >
        >
        > duh. Because 2+2=4?
        >
        More importantly 2x2=4

        > This is interesting. When I did this problem before, I always used
        > rationals for addends. Rational addends close to e always gave the
        > best results. So when Peter used natural numbers for addends, I
        > thought they must be 3's and 2's because they are closest to e. Now
        > you are showing that using a 4 can give the same result. Four is
        not very close to e.

        The addends must be 2 or 3, not because they are closer to e, but
        because an alternative to the addend 5, is the addends 2 and 3 which
        have a product of 6, which is greater than 5
        similarly 4 x 2 > 6, but 3x3 is even bigger.
        4x3 >7
        3x3x2>8
        3x3x3>9
        5x5 >10 and 3x3x4 is even bigger, because I showed above you would
        not use the addend 5.
        6x5>11, but you can do better than a 6 and a 5.
        etc

        so all addends greater than 4 would never be used.
        4 only gets a run because 2x2=4
        by using 4 you don't get a higher product, merely a smaller number
        [count] of addends.

        Peter
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