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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