Does there exist $n$ such that all numbers $n,2n,\dots,2000n$ have the same digits?

Question

Does there exist a number $n$ such that all numbers $n, 2n, 3n, 4n, \dots, 2000n$ have the same multi-sets of digits except zeroes?

(Having the same multi-sets of digits excepts zeroes means having equal number of ones, twos, ... , nines in the decimal expansion.)

A related question was already asked on MathSE, but the answers there does not provide an approach suitable for bigger numbers like 2000.

Answer 1

Suppose $N>2000$ is an integer such that the period length of the (eventually) repeating $\frac1N$ equals $N$. Then in computing the decimal expansion all remainders $1,\ldots,N-1$ occur at some place. Then the fractions $\frac1N,\frac2N,\ldots, \frac{2000}N$ turn out to lead to the very same period, merely shifted. In this situation, we have $\frac1N=\frac n{10^{N-1}-1}$ for some $1\le n< 10^N-1$ and conclude that $n,2n,\ldots,2000n$ indeed are obtained by rotating the digit sequences suitably (taking an appropriate number of leading zeroes into account).

(Actually, it would be sufficient that the remainders $1,2,\ldots,2000$ occur during the computation of the period, so the period length might be smaller than $N$.)

The question is: Do such $N$ exist? Primes are good candidates (for any other $N$, the order of $10$ cannot exceed $\phi(N)$). So for which primes $N$ is $10$ of order $N-1$? One such prime is $2017$ and that is $>2000$, thius solving the concrete problem - or rather $$ n=\frac{10^{2016}-1}{2017}=\underbrace{4957858205\ldots233019335647}_{2013\text{ digits}} $$ is. Intriguingly, each of the digits $1,2,4,5,7,8$ occurs $202$ times, each of $3,6,9$ occurs $201$ times in that $n$. In fact, it turns out that $2017$ is the smallest prime $>2000$ with this property. As additionally, $\phi(n)<2000$ for all composites $<2017$, we see that $2017$ is the smallest $N$ for which the above construction works. However, this does not completely rule out that smaller $n$ exist (where the $kn$ only "accidentally" have the same digit statistics).

See also sequence Full reptend primes in OEIS.

Answer 2

The answer is positive. Moreover, there exists a number $n$ such that additionally all numbers $2n,3n,\dots,2000n$ are obtained from $n$ by cyclic permutation of digits (naturally, we need to add some zeroes before $n$ for that). Here is this number:

0004330879168471199653529666522304027717626678215677782589865742745777392810740580337808575140753572975313988739714161974880900822867042009527934170636639237765266349068860978778692074491121697704634040710264183629276743178865309657860545690775227371156344737981810307492420961455175400606323083585967951494153313122563880467734950194889562581203984408834993503681247293200519705500216543958423559982676483326115201385881333910783889129493287137288869640537029016890428757037678648765699436985708098744045041143352100476396708531831961888263317453443048938934603724556084885231702035513209181463837158943265482893027284538761368557817236899090515374621048072758770030316154179298397574707665656128194023386747509744478129060199220441749675184062364660025985275010827197921177999133824166305760069294066695539194456474664356864443482026851450844521437851883932438284971849285404937202252057167605023819835426591598094413165872672152446946730186227804244261585101775660459073191857947163274144651364226938068427890861844954525768731052403637938501515807708964919878735383282806409701169337375487223906453009961022087483759203118233001299263750541359896058899956691208315288003464703334776959722823733217843222174101342572542226071892594196621914248592464270246860112602858380251190991771329579904720658293633607622347336509311390212213079255088783022953659592897358163707232568211346903421394543092247726288436552620181896925075790385448245993936769164140320485058466868774361195322650498051104374187960155911650064963187527067994802944997834560415764400173235166738847986141186660892161108705067128627111303594629709831095712429623213512343005630142919012559549588566478995236032914681680381117366825465569510610653962754439151147682979644867908185361628410567345171069727154612386314421827631009094846253789519272412299696838458207016024252923343438718059766132524902555218709398007795582503248159376353399740147249891728020788220008661758336942399307059333044608055435253356431355565179731485491554785621481160675617150281507145950627977479428323949761801645734084019055868341273278475530532698137721957557384148982243395409268081420528367258553486357730619315721091381550454742312689475963620614984841922910350801212646167171935902988306626245127760935469900389779125162407968817669987007362494586401039411

But how to find it?

Let us call a number $n$ $k$-cyclic if all numbers $2n,3n,\dots,kn$ are obtained from $n$ by cyclic permutation of digits. For example, $142857$ is $6$-cyclic: $2\cdot 142857 = 285714$, $3\cdot 142857 = 428571$, $4\cdot 142857 = 571428$, $5\cdot 142857 = 714285$, $6\cdot 142857 = 857142$.

The crucial step is the following observation (proof is available on request):

Let $p$ be a prime number such that $10$ is a primitive residue modulo $p$. Then the period of the fraction $1/p$ is $(p-1)$-cyclic.

For example, $10$ is primitive root modulo $7$, and from $1/7 = 0.(142857)$ we get that $142857$ is $6$-cyclic, which we saw already.

So we are left to find a prime number $p>2000$ such that $10$ is a primitive root modulo $p$. There is a useful Chebyshev theorem:

If $p=4q+1$ is prime, where $q$ is prime and $q = 2 \pmod 5$, then $10$ is a primitive root modulo $10$.

Now the smallest prime number greater than $2000$ and satisfying the Chebyshev theorem is $2309 = 4\cdot 577-1$. Therefore, the period of $1/2309 = 0.0004330879\dots$ is $2308$-cyclic (hence, $2000$-cyclic). This period is written above.

Answer 3

I told my friend Ruby how to check the numbers by Hagen, zhoraster and two test numbers by me (n=4212345, n=0). She did the check and got:

n digits:
[0:198, 1:202, 2:202, 3:201, 4:202, 5:202, 6:201, 7:202, 8:202, 9:201]
OK - congrats!

n digits:
[0:227, 1:231, 2:231, 3:231, 4:231, 5:231, 6:231, 7:231, 8:231, 9:230]
OK - congrats!

n digits:
[0:0, 1:1, 2:2, 3:1, 4:2, 5:1, 6:0, 7:0, 8:0, 9:0]
error:
n_2 digits:
[0:1, 1:0, 2:1, 3:0, 4:2, 5:0, 6:1, 7:0, 8:1, 9:1]

n digits:
[0:1, 1:0, 2:0, 3:0, 4:0, 5:0, 6:0, 7:0, 8:0, 9:0]
OK - congrats!

We then did a little search for numbers of the form $(10^{k-1}-1)/k$ for $k$ from $1$ to $3000$. See the log.