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From: Frederick Williams <freddywilli...@btinternet.com>
Newsgroups: alt.math.recreational
Subject: Re: Jacobian problem
Date: Fri, 02 Nov 2012 22:47:40 +0000
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Barry Schwarz wrote:
> 
> On Wed, 31 Oct 2012 23:57:36 +0000, Nick <Nick.S...@yahoo.co.uk>
> wrote:
> 
> >On 30/10/2012 21:29, Frederick Williams wrote:
> >> While looking for something else, I came across...
> >>
> >> "...a challenge from Jacob Bronowski (1908--1974): find the least
> >> integer (in base 10) such that moving the leading digit to the rear
> >> produces a new integer one and a half times the original. This puzzle
> >> carried the warning that computation might prove lengthy, and, indeed,
> >> the answer runs to 16 digits."
> >>
> >> (http://www-history.mcs.st-andrews.ac.uk/Extras/Bronowski_Gazette.html)
> >>
> >> The answer, and others with multiples other than one and a half, may be
> >> well-known.
> >>
> >> I hope the awful pun will be excused.
> >>
> >
> >4705882352941176
> >
> >and no.
> 
> 1176470588235294, 2352941176470588, and 3529411764705882 all satisfy
> the condition and are smaller.
> 
> And the computation is trivial.  If we represent the original number
> as x*10^n+y with 1<=x<=9, then the transposed number is 10y + x.  The
> problem is then reduced to finding n, x, and y such that
>     10y + x = 1.5(x*10^n + y)
> which reduces to
>     y = (3 * 10^n - 2) * x / 17
> Performing the arithmetic in C using unsigned long long (up to 20
> digits on my system) and varying x from 1 to 9 and n from 1 to 19 

I'm probably being thick but since y is an integer 17 must divide
(3 * 10^n - 2) * x, and since it's a prime it either has to divide
(3 * 10^n - 2) or x.  x is too small and (3 * 10^n - 2) is even...

> (less than 200 test cases) produced ten potential solutions.  Five
> were bogus because integer division truncation produced an odd y which
> can never satisfy the original condition.  The five valid solutions
> (including 5882352941176470) all had 16 digits.  I was a little
> surprised that there were no 17-20 digit solutions.