# Proof of n/7, n=1,2,3,4,5,6

1. Jun 4, 2004

### imathgeek

Hi there,

There was an interesting problem proposed to me by some office mates a couple of days ago: "Prove that n/7, where n= 1, 2, 3, 4, 5, 6, is a repeating decimal where the digits repeat in a cyclical manner." I presented a more general proof of any fraction where the base is not 2^x*5^y will be repeating and non-terminating. However, the proof presented did not satisfy their cyclical requirement. Any suggestions on explaining to them that my proof is satisfactory or what should I add to a proof that would show the cyclical nature of all repeating decimals?

Thanks

Ken

2. Jun 4, 2004

### arildno

It is not true that all repeating decimals are cyclic in the way that the n/7-set is cyclic

For example,
1/9=0.11...
2/9=0.22..

There is no "cyclic" connection between 1/9 and 2/9!

What you should do, is simply to perform the divisions 1/7,2/7,3/7,4/7,5/7,6/7
and identify the cyclic connection between these decimal expansions.

3. Jun 4, 2004

### Gokul43201

Staff Emeritus
If the decimal representation of p/q has repeating blocks of 'n' digits, then
(p/q)*(10^n -1) =k, an integer.

In the case of n/7, the repeating block has 6 digits. So all you need is for 999999 to be divisible by 7....and it is !

From this you can see that if q is even you won't have repeating blocks. Also, no 999...9 will be divisible by 5. So if q=5 there will not be repeating blocks. And since 9 is divisible by both 3 and 9, the repeating block for those denominators will only be 1 digit long. This, I imagine does part of the job of your general proof for q=2^x*5^y.

i.e: For q<10, repeating blocks will seen only if q=3,7,9, and the size of the repeating blocks are 1,6,1 respectively.

Last edited: Jun 4, 2004
4. Jun 4, 2004

### Gokul43201

Staff Emeritus
Oops...didn't realize that you were supposed to show that the all numbers of the form n/7 had cyclical repetitions of the SAME block of digits.

Hmmm... it doesn't really say this anywhere in your post...

5. Jun 4, 2004

### arildno

Well that's how I interpreted "in a cyclical manner" (since I happen to know about this curious property of the n/7-fractions)

However, it seems you have a lot more general knowledge than me about number theory;
so I'd like to ask you:
What conditions must be present in order for the fractions m/n (1<=m<n) represent a "cyclic" permutation of the same block of digits?

I have a sneaking suspicion that n=7 is the only number where we have this property, but I'm not sure..

6. Jun 4, 2004

### arildno

Just for the record, we have:
1/7=0.142857..
2/7=0.285714..
3/7=0.428571..
4/7=0.571428..
5/7=0.714285..
6/7=0.857142..

7. Jun 4, 2004

### Gokul43201

Staff Emeritus
I have a vanishingly small knowledge of number theory, and even that little bit is self taught...but I shall see where this goes...

It's not hard to show that n/q and m/q can be written with repeating blocks (of size k) of the same numbers if there exists some r<k satisfying
m*10^r == n (mod q) for all 0<n<m<q. {where a==b (mod c) means a-b is divisible by c}

If your suspicion is right, r's can be found only for q=7.

(I actually wrote the proof down till this point...and then I accidentally hit some button and lost everything...will do it again if required.)

If we restrict q to [1..10], then it's easy to eliminate q=2,3,4,5,6,8,9,10.

For example, take q=3 : 10==1 (mod 3) => 10^r == 1 (mod 3) for all r => m*10^r==m (mod 3).
So if we pick some (m,n) with m<>n, then there will be no r that satisfies the required condition.

Finding all q's in Z for which r's can be found seems harder. It is easy to show that q must be odd. Further, q will need to be prime for n/q to have a decimal representation with repeating blocks.

But for now, this is as far as I will try to go.

Last edited: Jun 4, 2004
8. Jun 4, 2004

### Gokul43201

Staff Emeritus
Also, the above condition can be shown to be satisfied by q=7. There are only 36 pairs of (m,n) to check and clearly the condition for (m,m) is satisfied by r=6, so there's ONLY 30 pairs to go !

So that answers imathgeek's original question. Perhaps his general proof is the way to go. That would be nice to see.

I'm sure there's a nicer way to prove this.

Last edited: Jun 4, 2004
9. Jun 4, 2004

### Gokul43201

Staff Emeritus
Uh Oh...check out 17,47.

10. Jun 4, 2004

### arildno

Cool, I'll ponder on this for a while..

11. Jun 4, 2004

### imathgeek

oops.

yeah, you're right. I wasn't thinking too well before that first cup of coffee. Now, i am back to the proof.

Pure math is fun, but I still enjoy applied math. :)

Have fun,

Ken

Last edited: Jun 4, 2004
12. Jun 4, 2004

### imathgeek

I have done that. Thanks for the tip, though.

13. Jun 4, 2004

### CTS

14. Jun 4, 2004

### Gokul43201

Staff Emeritus
"No general method is known for finding full reptend primes. "

Hmmm...a few wasted minutes, but I did enjoy them.