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June 5th, 2014, 01:28 PM   #1
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repeating digits in decimal expansion of 1/3^n is 3^(n-2)

Hi

Is there any easy way to do this problem.

Show that for n>=2, the number of digits in the block of repeating digits in the decimal expansion of 1/3^n is 3^(n-2)

Thanks in advance and Regards
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June 5th, 2014, 03:16 PM   #2
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Induction perhaps.
For n = 2, we have 1/3^2 = 0.11...............
Which has period 1 = 3^(2-2) and the sum of the digits of the period isn't divisible by 3.

Suppose, we have the number 1/3^k for k >= 2, which has a period of repeating decimals of 3^(k-2), and the sum of the digits in a period don't divide 3, then for 1/3^(k+1), we can concatenate 3 periods of 1/3^k such that the sum of the result is divisible by 3. Than we divide the result by 3 and we get the period of 1/3^(k+1) and 3^(k-1) is the smallest period to be found, which completes the proof.

P.S. For 1/27, we have 037037037, for which we include the first digit 0 such that 037037037/3 = 012345679, which is the repeating decimals of 1/81.
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June 5th, 2014, 07:06 PM   #3
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Hi Hoempa

I don't understand clearly.

1. What do you mean by "Sum of the digits in a period don't divide 3"

2.1/3^(k+1), how can we concatenate 3 periods of 1/3^k which makes it divisible by 3?

Thanks for your help though. I expect little bit easier version.

Thanks
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June 6th, 2014, 12:26 AM   #4
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1.)
The period of these digits for 1/9 is 1. The sum of these digits is 1, which doesn't divide 3. So the sum of the digits in the period doesn't divide 3.

The period of these digits for 1/27 is 037. The sum of these digits is 0+3+7=10, which doesn't divide 3. So the sum of the digits in the period doesn't divide 3. And so on.

2.) from the repeating decimals of 1/27, 037, we can find the repeating decimals for 1/81, by concatenation of 3 periods, 037, 037 and 037. This gives
037037037. The sum of the result is divisible by 3; it's 3 * (digitsum of 037).
The decimal period of 1/81 is 037037037/3 = 012345679. See?
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June 6th, 2014, 08:16 AM   #5
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Hi Hoempa

Yes...it is ok but we started our assumption with the proof already. Then we showed that it persists in any power of k.

Is there any straight way to do this?

However thanks
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June 8th, 2014, 10:42 AM   #6
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I used, informally perhaps, induction. Are you familiar with that?
Here is a brief example:
Suppose, theoretically, that if the sun shines on one day, the sun also shines the next day. Today, the sun shines. Is there any day after today the sun doesn't?
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June 8th, 2014, 10:48 AM   #7
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Quote:
Originally Posted by Hoempa View Post
Suppose, theoretically,...
You mean: Suppose, theatrically
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June 8th, 2014, 11:30 AM   #8
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Hm, yeah, nice how these two interfere
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June 8th, 2014, 12:56 PM   #9
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Hi Hoempa

No i am not familiar with induction process.

But i understand your logic but still this is informal approach. Kind of abstract math.

Thanks
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June 8th, 2014, 01:48 PM   #10
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Okay, you can read more about it here:Mathematical induction - Wikipedia, the free encyclopedia
The approach is self is formal, maybe work on it is somewhat informal written. You could submit such a proof as a good answer. Sorry, I don't know another way to do it.
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