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Re: No Putnam spoilers please
Posted:
Dec 14, 2012 6:05 AM
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On Dec 13, 2:01 pm, Dr J R Stockton
<reply1...@merlyn.demon.co.uk.invalid> wrote:
> In sci.math message <MPG.2b2efda4be25e802989...@news.eternal-
> september.org>, Sun, 9 Dec 2012 19:58:25, Wasell
> <Was...@example.invalid> posted:
>
> >On Sat, 8 Dec 2012 19:13:02 +0000, in article
> ><1tj7wWW+E5wQF...@invalid.uk.co.demon.merlyn.invalid>, Dr J R Stockton
> >wrote:
>
> >> Can anyone (or more) please provide here the last ten decimal digits
> >(in
> >> order) of ((3^349)-1)/2, freshly and independently calculated and not
> >> copied from any other medium, and not using my LongCalc or VastCalc?
>
> >$ perl -Mbigint -e 'print substr( ((3**349)-1)/2, -10 )'
> >7379284041
>
> Real Life intervened. Thanks to all four. I've not checked the whole
> of their long numbers, but parts are right so no doubt about the rest.
> I had also used Richard's approach, in JavaScript.
>
> Some years ago, while reading a book that showed (roughly)
>
> (3^349-1)/2
> =
> three lines of digits
> =
> two lines of digits
> *
> two lines of digits
>
> I noticed a missing digit near the beginning of the second "two lines",
> and reported it. I was evidently checking the expression against the
> multiplication. I have now noticed that your second 7 is, in the book,
> an 8 - 7389284041 - and wanted a cross-check before reporting the new
> fault. OTOH, before doing that, I should re-check ALL digits.
>
(3^349-1)/2 =
94042850889984510998289152320438541798532018021653956283741193211654025280185459.P87
Where P87 indicates a prime cofactor of 87 digits.
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