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Jacobian problem
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Frederick Williams
Oct 30, 2012, 5:29:33 PM10/30/12
to
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.
--
I have seen elephants paint very competently... but not in Cumbria.
"...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.
--
I have seen elephants paint very competently... but not in Cumbria.
Nick
Oct 31, 2012, 7:57:36 PM10/31/12
to
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
> 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.
>
and no.
Christopher J. Henrich
Nov 1, 2012, 2:08:17 PM11/1/12
to
In article <5091baee$0$12366$c3e8da3$1920...@news.astraweb.com>, Nick
[spaces added:]
> 4705 8823 5294 1176
>
Try this:
1176 4705 8823 5294
And note that the groups of four digits have been cyclically permuted.
Spooky?
--
Chris Henrich
http://www.mathinteract.com
We've tamed the lightning and taught sand to give error messages. - Nancy
Lebovitz
> 4705 8823 5294 1176
>
Try this:
1176 4705 8823 5294
And note that the groups of four digits have been cyclically permuted.
Spooky?
--
Chris Henrich
http://www.mathinteract.com
We've tamed the lightning and taught sand to give error messages. - Nancy
Lebovitz
Nick
Nov 1, 2012, 2:50:49 PM11/1/12
to
On 01/11/2012 18:08, Christopher J. Henrich wrote:
> In article <5091baee$0$12366$c3e8da3$1920...@news.astraweb.com>, Nick
> <Nick...@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.
>>>
>>
> [spaces added:]
>> 4705 8823 5294 1176
>>
> Try this:
> 1176 4705 8823 5294
>
> And note that the groups of four digits have been cyclically permuted.
> Spooky?
>
I was going to say I hadn't quite got straight in my head what I was
> In article <5091baee$0$12366$c3e8da3$1920...@news.astraweb.com>, Nick
> <Nick...@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.
>>>
>>
> [spaces added:]
>> 4705 8823 5294 1176
>>
> Try this:
> 1176 4705 8823 5294
>
> And note that the groups of four digits have been cyclically permuted.
> Spooky?
>
doing for the 117 case but it appears I didn't have it straight for
other cases too, probably I still haven't.
2352941176470588
3529411764705882
5882352941176470
again permutations. That's more than spooky
Barry Schwarz
Nov 2, 2012, 2:16:53 AM11/2/12
to
On Wed, 31 Oct 2012 23:57:36 +0000, Nick <Nick...@yahoo.co.uk>
wrote:
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
(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.
--
Remove del for email
wrote:
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
(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.
--
Remove del for email
Nick
Nov 2, 2012, 10:37:18 AM11/2/12
to
So surely all we are checking is that (3 * 10^n - 2) is divisible by 17,
as x and 17 are co-prime.
If it is divisible the solutions are just the multiples of (3 * 10^n -
2)/17 which do not add a digit. i.e. x=1,2,3,4,5.
The permutations bit seems to be related to the fact 3/17 is a recurring
decimal of length 16.
The next valid solutions are when n = 31 giving a 32 digit number.
Frederick Williams
Nov 2, 2012, 6:47:40 PM11/2/12
to
(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...
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