## Re: Problem 320

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• ... 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
--- 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
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
>
• ... 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
--- 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