Permutation of digits by multiplication

Here is an expository write-up of this post.

The number N=142857 is interesting because of the following property

\begin{aligned}N&=142857\\ 2N&=285714\\ 3N&=428571\\ 4N&=571428\\ 5N&=714285\\ 6N&=857142\end{aligned}

and 7N=999999. The last equation means that

\displaystyle\frac 17=0.\overline{142857}.

So multiplying N by 1,2,3,4,5,6 simply permutes the digits of N. As elements of S_6 these permutations correspond to (using cycle notation)

\hbox{id}, (153)(264), (165432), (135)(246), (123456), (14)(25)(36)

respectively. We can read off the following from this list.

  • Since (165432) and (123456) are full-length cycles, 3 and 5 are primitive roots modulo 7.
  • 2 and 4 have order 3 modulo 7, while 6 has order 2.

In general, if p is an odd prime and b is a positive integer with order d modulo p, then 1/p will have period (b^d-1)/p of length d in base b. This follows simply from observing that if b^d-1=mp then

\displaystyle\frac 1p=\frac{m}{b^d-1}=\frac{m}{b^d}+\frac{m}{b^{2d}}+\frac{m}{b^{3d}}+\cdots.

Now let b>p be a primitive root modulo p. We shall work in base b. Then 1/p will have period L=(b^{p-1}-1)/p of length p-1. Since the periods of 1/p,2/p,\dots,(p-1)/p are L,2L,\dots,(p-1)L, all of length p-1, which are all powers of some permutation of the digits of L, and no two of which are congruent modulo b, it follows that the digits of L are all distinct. So we’ve shown the following.

Theorem. If p is an odd prime and b>p is a primitive root modulo p, then the base b expansion of 1/p has period of length p-1 with distinct digits.

Taking p=7, b=10 gives our previous example. So the prime 7 in base 10 is indeed very special, since it is the only prime less than 10 for which 10 is a primitive root.


1 Comment

Filed under Number theory

One response to “Permutation of digits by multiplication

  1. it seems 8N=1142856, it can be expressed like N=142857…..and the sum of the first and the last digit of 8N is the last digit of N. the digits in the middle of these two are the same to the digits except the last of N, 14285……….. again , 2N= 285714…….and 9N= 1285713 where, 1+3=4=the last digit of 2N…….the middle is same, 28571… goes on till 14N. then again 15N=2142855……N=142857……5+2=7…….it also goes on till 21N……………i want to say that it comes after every 7 turns…… at 1st turn (8N) the first digit is 1 and the last is the rest of 7……..then at second turn (15N), the first digit is 2 and the last is the rest of 7………then again……….1/7=0.142847 for the first…….it also 2/7=0.285714, 3/7=0.428571…………it goes on……..i think it will continue for 49N………….actually it’s nothing important but sounds interesting…………

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