# Regularities when $n$ and $2n$ contain the same digits

Suppose we would like to find positive integers $n$ such that the base-10 representations of $n$ and $2n$ contain precisely the same digits.

$142857$ is a well-known example, and computer search quickly produces many others, of which the smallest is $125874$. In fact, there are 12 solutions of 6 digits, all of which are permutations of $125874$; if one allows leading zeroes, there are 36 other solutions less than 1000000, all permutations of $0142857$. (For example, the smallest such is $0512874$.)

There are larger solutions that are a bit different, but they all seem to contain the digits 1, 4, 2, 8, 5, and 7 exactly once each. (Computer search verifies this for $n$ up to $63.6\cdot 10^6$, by which point there have been $6000$ solutions, each one containing 1,4,2,8,5, and 7 exactly once each.) Obviously there are larger solutions that contain each of 1, 4, 2, 8, 5, and 7 exactly twice each, obtained by concatenating two smaller solutions, but the pattern of $124578$ appearing as a group seems to persist. For example, the largest solution less than $10^8$ is $09875421$.

In other bases, there are much simpler solutions. For example $52_8 = 2\cdot 25_8$.

My overall question is:

What is going on here?

But in case that is too vague, here are some more specific questions:

1. Although I don't know an argument offhand, it's easy to imagine that there could be an algebraic argument showing that $142857$ has this property, since $142857$ is the useful number $\frac1n{\bigl(10^{n-1}-1\bigr)}$ for $n=7$. But why should all the other small solutions be permutations of this number? In particular, is there a way to deduce, without exhaustive search, that the minimal positive solution is $125874$? Or that any solution must contain the digits $1,2,4,5,7,$ and $8$, and so the minimal solution must be at least $124578$?

2. Other digits appear mixed into the $124857$ digits. For example, $12457899, 12356784,$ and $49712859$ are solutions. But none of $1,2,4,5,7,$ or $8$ seems to appear in a solution except a part of a group. Is it the case that every solution contains an equal number of each of these 6 digits?

3. Solutions in base 8 do not show so much regularity. The digits 3 and 6 always appear together, as do the digits 1 and 4. (5 and 2 are not closely coupled in this way; for example $1042_8\cdot 2 =2104_8$.) What is the difference between the base 8 and the base 10 situation? In base 16, $08124$ and similar numerals are solutions, which is suggestive, but I'm not sure of what.

I have a partial explanation, which I'll write up in more detail later on. I do have method that would allow someone to find a 6-digit base-10 solution in a reasonable amount of time with only pencil and paper; perhaps 30–60 minutes. (The difficulty is that there are no small solutions in base 10; it takes time to find this out, and then to perform the analysis for $n=6$.)

Meantime:

1. There are solutions in base 10 that do not include a group of 124578.
2. I can prove that smallest such has at least 10 digits.
3. An example is 4,225,561,128. The example 1,577,438,874 completely refutes amy theories about 124578 always appearing in groups.

I do not yet have any good explanation for why the 6 common digits are 124578, or whether the appearance of $\frac17(10^6-1)$ is a coincidence.

[ Added 2014-06-11: This is written up on the math.se blog. The short explanation of the behavior is that there are fairly strict conditions on the set of digits that can appear in a solution, but if these conditions are satisfied, the order of the digits is only weakly constrained, because the question only asks for a number where the digits of $2N$ are in some order. ]

Working modulo 9, we have your $\{1,2,4,8,7,5\}$ as $$2^0=1, 2^1=2, 2^2=4, 2^3=8, 2^4=7, 2^5=5, 2^6=1$$

I don't have a direct connection between this observation and your data, but I would conjecture that if we worked in base 12 rather than base 10, we would take mod $11$ and then all the digits would appear much more often.

• This seems related to the fact that 3 and 6, when they do appear, always appear together; we have $3\cdot 2 = 6$ and $6\cdot 2 = 3\pmod 9$. On the other hand, 3 and 6 also appear together in base 8, where we would not expect it, since $6\cdot 2\not\equiv 3\pmod 7$.
– MJD
Commented May 5, 2014 at 18:47