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        riddles >> easy >> base 3
        
(Message started by: Christine on Mar 31st, 2013, 12:26pm)
Title: base 3
Post by Christine on Mar 31st, 2013, 12:26pm
I don't remember how to do this type of operation:

Repeated permutations of (0,1,2) (base 3)

5/26 = 0.012…,
7/26 =  0.021…,
11/26 = 0.102…,
15/26 = 0.120…,
19/26 = 0.201…,
21/26 =  0.210…

How do you find these rational numbers?
Title: Re: base 3
Post by Grimbal on Mar 31st, 2013, 1:16pm
26 is 27-1.  27 is 3^3 (3 because of the base, 3 because of 3 digits).
If you took 27, you would just have 3 digits.  Taking 26 makes it periodic.

5 is 123, 7 is 213, 11 is 2013, etc.
Title: Re: base 3
Post by Christine on Mar 31st, 2013, 10:53pm

on 03/31/13 at 13:16:12, Grimbal wrote:
26 is 27-1.  27 is 3^3 (3 because of the base, 3 because of 3 digits).
If you took 27, you would just have 3 digits.  Taking 26 makes it periodic.

5 is 123, 7 is 213, 11 is 2013, etc.


It's still not clear to me.
I'll take another example: Let's find fractions 0. ...
so that we get permutations of {0,1,2,3}

75 -> 10234, 99 -> 12034, 108 -> 12304, 78 -> 10324,
135 -> 20134, 147 -> 21034, etc.

what are these rational numbers?
Title: Re: base 3
Post by towr on Mar 31st, 2013, 11:12pm
1/(bn-1) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifi=1..inf (1/bn)i

1/(10-1) = 1/9 = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifi=1..inf 0.1i = 0.111...
1/(100-1) = 1/99 = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifi=1..inf 0.01i = 0.010101...
1/(1000-1) = 1/999 = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifi=1..inf 0.001i = 0.001001001...

1/(3-1) = 1/2 =  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifi=1..inf 0.13i = 0.111...3
1/(9-1) = 1/8 =  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifi=1..inf 0.013i = 0.010101...3
1/(27-1) = 1/26 =  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifi=1..inf 0.013i = 0.001001001...3

1/(4-1) = 1/3 =  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifi=1..inf 0.14i = 0.111...4
1/(16-1) = 1/15 =  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifi=1..inf 0.014i = 0.010101...4
1/(64-1) = 1/63 =  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifi=1..inf 0.0014i = 0.001001001...4

So to repeat k digits x..z in base b, take 1/(bk-1) * x..z
And of course you can shift it by dividing by bm and add some not repeating term if you want.

So, e.g.
0.1023...4
= 10234 * http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifi=1..inf 0.00014i
= 10234 * 1/(44-1)
= 10234 * 1/(256-1)
= 75/255
= 15/51
Title: Re: base 3
Post by Grimbal on Apr 1st, 2013, 6:06am

on 03/31/13 at 22:53:17, Christine wrote:
75 -> 10234, 99 -> 12034, 108 -> 12304, 78 -> 10324,
135 -> 20134, 147 -> 21034, etc.

what are these rational numbers?

Short answer: the denominator is 44-1 = 255.
75/255 = 0.294117647058824 = 0.10231023...4
99/255 = 0.12031203...4

As towr mentioned, the fraction needs to be simplified.
Title: Re: base 3
Post by Christine on Apr 1st, 2013, 9:59am
Thanks Grimbal and towr. I understand better now
Title: Re: base 3
Post by whizen on May 30th, 2013, 2:25pm
Loved this question and the answers.

Christine...
Where do you get all these questions? ( I have seen some more interesting ones from you in the forum )


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