7

It is easy to verify that $$\frac 17=\frac {142857}{999999}$$ where $142857$ is the decimal period of $\frac 17$.

This period, which has six different digits, has the property that when multiplied by $1,2,3,4,5,6$, the respective products have the same six different digits in different position (Clearly, multiplied by $7$ must give $999999$).

Is $ 142857 $ the only six-digit number that has this property?

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Piquito
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    Thus will be true for the period of $1/n$ iff the order of $10 \bmod n$ is $6$. – lhf Dec 17 '15 at 15:55
  • I can't find the post I was looking for, but here's something similar. https://math.stackexchange.com/questions/56989/cyclic-numbers-are-characterized-by-the-reciprocals-of-full-reptend-primes – pjs36 Dec 17 '15 at 16:03
  • See https://en.wikipedia.org/wiki/Cyclic_number,, http://math.stackexchange.com/questions/443/why-is-the-decimal-representation-of-frac17-cyclical and http://math.stackexchange.com/questions/56989/cyclic-numbers-are-characterized-by-the-reciprocals-of-full-reptend-primes – lab bhattacharjee Dec 17 '15 at 16:24

1 Answers1

6

Thus will be true for the period of $1/n$ iff the order of $10 \bmod n$ is $6$, but you'll have to consider different sets of multipliers.

For $n<100$, the examples are $n=7, 13, 21, 39, 63, 77, 91, 97$. (*)

For $n=13$, the number is $076923$ (let's accept this as having six digits) and there are two cycles: one for the multipliers $1,3,4,9,10,12$ and one for $2,5,6,7,8,11$.

(*) Apparently, there are only $53$ examples; see A059892 and A226477.

lhf
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  • I remarked this property trying to solve the problem in http://math.stackexchange.com/questions/1578471/prove-that-if-7n-3n-is-divisible-by-n1-then-n-must-be-even. The example $\frac{1}{17}$ given in the comment of pjs36 is really good and answer to a generalization of this question (however I do not really like the arrangement with 1/13). Thank you. – Piquito Dec 17 '15 at 19:47