9
$\begingroup$

Let $X:=\{$ positive integers that contain the digit $2\}$

For fixed $m,n\in\mathbb{N},$ define the A.P. $S_{m,n:=}\ \{m,\ m+n,\ m+2n, \ldots\}\ .$

I am interested in $S_{m,n}\cap X,$ and $S_{m,n}\cap (\mathbb{N}\setminus X).$

It is clear that $ S_{2,10}\subset X $ and so $\vert S_{2,10}\cap X\vert = \infty.$

Next, observe that $X$ has natural density $1$: In order for a number with $10^6$ digits to have no $2'$s, the first digit must not be $2,$ ($8$ out of $9$ ways), and the next $999999$ digits must not be $2,$ ($9$ out of $10$ ways for each of the $999,999$ digits) - so $8*9^{999999}$ out of $9*{10}^{999999}.\quad \frac{8*9^{999999}}{ 9*{10}^{999999}}= \left(\frac{8}{9}\right)\left( \frac{9}{10} \right)^{999999} \approx 0.$ So relatively speaking, there are very few numbers with no digit $2.$

The question:

As already mentioned, there is a subset of $X$ that is an infinitely long A.P, for example, $S_{2,10}.$ Is there a non-trivial infinitely long A.P., $S_{m,n}$, that is also a subset of $X?$ $S_{2,10}$ is "trivial" because the last digit of all members of this A.P. is $2.$ Non-trivial, therefore, means that the $k-$th - last digit of every member of $S_{m,n},$ is not $2,$ for every $k.$ In other words, $2$ appears in every member of $S_{m,n},$ but it's place value keeps changing (more precisely, it isn't in the same position in every member of $X$.)

In all of the above, the number $2$ was arbitrary, and the answer is probably the same if we replace $2$ with any other digit.

I suspect the answer is yes, but that such a number will have a large amount of digits. On the other hand, the answer may be "no" due to some application of, for example, Dirichlet's approximation theorem.

$\endgroup$
6
  • 2
    $\begingroup$ One may assume without loss of generality that $n$ is not a multiple of $10$. If it is, then either the last digit of $m$ is $2$ (in which case $S_{m,n}$ is a "trivial" subset of $X$), or else the last digit of $m$ is not $2$ (in which case one could remove it from $m$ to get a new number $m'$ for which $S_{m',n/10}$ is still a subset of $X$). $\endgroup$
    – Geoffrey Trang
    Commented Jul 6 at 14:48
  • $\begingroup$ @GeoffreyTrang good thinking. $\endgroup$
    – Adam Rubinson
    Commented Jul 6 at 15:46
  • $\begingroup$ So the last digit is not $0.$ $\endgroup$
    – Adam Rubinson
    Commented Jul 6 at 16:00
  • $\begingroup$ I'm guessing that $x_{n+1}=x_n+10$ for $x_0=2$ is trivial by the standards of the question. $\endgroup$
    – controlgroup
    Commented Jul 13 at 14:43
  • $\begingroup$ @controlgroup Yes, I specifically rule out this case. I suggest you read the body of the question carefully $\endgroup$
    – Adam Rubinson
    Commented Jul 13 at 14:49

1 Answer 1

Reset to default
3
+50
$\begingroup$

Every arithmetic sequence, in the nontrivial sense described by OP, contains a number whose digits are all $0$ or $1$.

I will show, if there is an $S_{m,n}\subset X$, then we may assume $n$ is coprime to $10$. Then, take a subsequence, assume that $n=10^k-1$. Then spread out all the digits of $m$, find an $m'$ in the sequence with all digits $0$ or $1$.

As noted by @GeoffreyTrang, we can assume $n$ is not a multiple of $10$.

If $n$ ends in $5$, then either $S_{m,2n}$ or $S_{m+n,2n}$ don't end in $2$. So we can find $m'$ with $S_{m',n/5}\subset X$. So we can assume $n$ is not a multiple of $5$.

Similarly we can assume $n$ is not a multiple of $2$.

Consider the remainders of powers of $10\pmod n$ Two of them must be equal, say $10^j=10^{j+k}\pmod n$. $n$ is coprime to $10^j$ so there is a $p$ with $np=10^k-1$. By taking a subsequence of our original $S$, let the new $n'=10^k-1$.

Write $$m=\sum 10^{a_i}\\ eg. 23=10+10+1+1+1$$ For example, if $n'=999$, then $20=10010\pmod{999}$ and $3=1001001\pmod{999}$. So $23=1011011\pmod{999}$. In this way, any number $m$ is equivalent $\pmod{10^k-1}$ to a number that contains only $0$s and $1$s. This number is in the sequence.
This contradiction shows that no nontrivial $S_{m,n}$ is a subset of $X$.

$\endgroup$
4
  • $\begingroup$ I don't understand "If $n$ ends in 5, then either $𝑆_{𝑚,2𝑛}$ or $𝑆_{𝑚+𝑛,2𝑛}$ don't end in 2. So we can find $m'$ with $S_{m',n/5} \subset X$." Why is it enough to check that the last digit of the elements in these sets isn't 2? Don't you need something much stronger to conclude that the spacing $n$ between elements in $S_{m,n}$ can be reduced to $n/5$ (as it is in $S_{m',n/5}$)? $\endgroup$
    – Erik T
    Commented Jul 15 at 3:21
  • 1
    $\begingroup$ We have two sub-sequences $S_{m,2n}$ and $S_{m+n,2n}$. $2n$ is a multiple of $10$ so we can chop off the final digit of both $m'$ and $2n$, like @GeoffreyTrang did, leaving a new sequence with $2n/10=n/5$. Then keep dividing $n$ by $5$, and adjusting $m'$, until $n$ is no longer a multiple of $5$. $\endgroup$
    – Empy2
    Commented Jul 15 at 4:13
  • $\begingroup$ ah, I see. That's clever. $\endgroup$
    – Erik T
    Commented Jul 15 at 4:15
  • $\begingroup$ Took me awhile to understand this one, but I finally got it this afternoon. Key step I didn't get: to make sure the higher order terms are all 0 or 1 we add every higher values of $k$ in $10^k + 1$. Also, pick $k$ sufficiently large, so that $m$ has (perhaps several) fewer digits than $10^k$. $\endgroup$
    – Artimis Fowl
    Commented Jul 15 at 21:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .