From Newsgroup: rec.puzzles

This puzzle is taken from 'Amusements in Mathematics' by Henry Ernest
Dudeney.

If we number six cards 1, 2, 4, 5, 7 and 8, and arrange them on the
table in this order:-

1 4 2 8 5 7

we can demonstrate that in order to multiply by 3 all that is necessary
is to remove the 1 to the other end of the row, and the thing is done.
The answer is 428571. Can you find a number that, when multiplied by
three and divided by 2, the answer will be the same as if we removed the
first card (which in this case is to be a 3) from the beginning of the
row to the end?
--
David Entwistle

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From Newsgroup: rec.puzzles

It happens that David Entwistle formulated :

1 4 2 8 5 7

maybe this can help?

https://en.wikipedia.org/wiki/Cyclic_number
--
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From Newsgroup: rec.puzzles


David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> posted:

Problem repeated as soiler warning space.

This puzzle is taken from 'Amusements in Mathematics' by Henry Ernest Dudeney.

If we number six cards 1, 2, 4, 5, 7 and 8, and arrange them on the
table in this order:-

1 4 2 8 5 7

we can demonstrate that in order to multiply by 3 all that is necessary
is to remove the 1 to the other end of the row, and the thing is done.
The answer is 428571. Can you find a number that, when multiplied by
three and divided by 2, the answer will be the same as if we removed the first card (which in this case is to be a 3) from the beginning of the
row to the end?

3529411764705882

With a little manipulation I found that I needed an N where
(10^N - 11) was a multiple of 17.

Cheers,
James
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From Newsgroup: rec.puzzles

In article <1757698508-4353@newsgrouper.org>,
James Dow Allen <user4353@newsgrouper.org.invalid> wrote:

(10^N - 11) was a multiple of 17.

10^N - 1, I think.

-- Richard
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From Newsgroup: rec.puzzles


richard@cogsci.ed.ac.uk (Richard Tobin) posted:

In article <1757698508-4353@newsgrouper.org>,
James Dow Allen <user4353@newsgrouper.org.invalid> wrote:

3529411764705882

With a little manipulation I found that I needed an N where
(10^N - 11) was a multiple of 17.

10^N - 1, I think.

-- Richard

You think, but did you do the exercise?
Remember that "cyclic numbers" cycle when multiplying by integers,
but here we are multiplying by 1.5.

I do NOT know the theory of such cycles, but will give
the boring details of my calculation.

Boring Boring BORING. You've been warned.

It is easy to see that the solution will have the form
(3*10^N + A)*3 = 2*(10A + 3)
This leads directly to
9*10^N = 17A + 6
The left side is a multiple of 9. The substitution
A = 9B + 6
ensures that the right side is also a multiple of 9.
Simplifying we find that we need
(10^N - 12) to be a multiple of 17. Not (10^N - 1)

But not (10^N - 11) that I claimed above either!! What gives?

Enter afore-mentioned "With a little manipulation."
It was easy to see that the target number must end in 2.
A little more effort leads to 82 as the two final digits.
With that starting point, a derivation as above does indeed lead
to the (10^N - 11) target (with an "off by 2" on the resultant N).
I suppose other manipulations might lead to (10^N - 1).

My solving for 2 digits before engaging my brain turned out
to be serendipitous! At the moment I have no Unix or cygwin
machine and haven't downloaded a 'bc' for Windows (if there even is
such a thing?) So I was using Google or wolframalpha as calculator.
But both of these seem to conk out at about 12 significant digits.
Enough to solve given the '82' finale, but not enough from scratch.

Cheers,
James
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From Newsgroup: rec.puzzles

In article <1757762204-4353@newsgrouper.org>,
James Dow Allen <user4353@newsgrouper.org.invalid> wrote:

10^N - 1, I think.

You think, but did you do the exercise?

Yes, but my N was not the same as yours.

Taking the full number as X, and the number of digits as N, so the
first digit 3 corresponds to 3 . 10^(N-1),

10 (X - 3.10^(N-1)) + 3 = 3X/2

20X - 6.10^N + 6 = 3X

17X = 6 (10^N - 1)

So we need a power of 10 that is one more than a multiple of 17. The
first is 10^16.

-- Richard
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