I have a following question:

Can we find for any natural number $n \in \mathbb{N}$, a sequence of only $\{0,1\}$ as elements such that the sequence has exactly $n\ 1's$ and is divisible by $n$ when viewed as a natural number. I can form easily a number with exactly $n$ ones but then how do I check that this sequence is divisible by $n$. Maybe the problem requires some combinatorial argument which I cant see properly.$\\$

For example, $3 \in \mathbb{N} $ corresponds to $111$ and $101010$ in the set of sequences. Both these are elements such that they have $3$ ones and $3$ divides them. If $n=4$, then we have the sequences $111100$ and $1101100$ etc. such that there are $4$ ones and $4$ divides them. Also, if $n=5$, we have $111110$, $10101110$ etc, which are divisible by $5$ and have $5$ ones. The point is that there can be more than one sequences which satisfy the given properties but how to show that one exists when the numbers are larger say $n=143$ etc. $\\$

Any hint/help would be nice. Thanks

  • 1
    $\begingroup$ I don't understand your description of your goal. Please edit the question to show us examples of sequences that work and that don't work for the numbers from $1$ to $10$. Then maybe we can help. $\endgroup$ – Ethan Bolker Apr 17 at 10:03
  • $\begingroup$ Not clear. Are you allowing infinite sequences of $1's$ and $0's$? All your examples are finite. If you restrict to finite one, then the list of such is countable. $\endgroup$ – lulu Apr 17 at 10:37
  • $\begingroup$ Oh yes they are countable sequences. $\endgroup$ – Rick Apr 17 at 10:39
  • $\begingroup$ Not following. You said "it is an uncountable set". Well, what is "it"? I assumed you meant the set of sequences of $0's, 1's$. But if that is only true if you allow infinite sequences of $0's, 1's$. Please clarify. $\endgroup$ – lulu Apr 17 at 10:47
  • $\begingroup$ ok, now I get that I made a mistake, thanks for pointing that out. Moreover, I was more concerned with the divisibility part/ how to show the existence of a number which is divisible by $n \in \mathbb{N}$. $\endgroup$ – Rick Apr 17 at 11:16

Your question is unclear, but perhaps the following observation will be useful.

Suppose $\gcd(n,10)=1$. Then there is some $P>0$ such that $10^P\equiv 1 \pmod n$. It follows that $10^0, 10^P, \cdots, 10^{(n-1)P}$ are all $\equiv 1 \pmod n$. But then $\sum_{i=0}^{n-1} 10^{iP}\equiv 0\pmod n$ so you have an example of the sort of sequence you want.

Of course, if $\gcd(n,10)>1$ we can write $n=2^a5^bN$ with $\gcd(N,10)=1$. Then perform the above construction using $n$ $1's$ to get a sequence divisible by $N$. After that multiply by $10^{\max(a,b)}$ to finish.

Note: this construction isn't exactly efficient, meaning that there is no suggestion that it finds a minimal example. If $n=7$, say, $P=6$ so the construction leads us to $1+10^6+10^{12}+10^{18}+10^{24}+10^{30}+10^{36}$, or $$1000001000001000001000001000001000001$$ while, of course, $11101111$ works as well.

  • $\begingroup$ Thanks very much for that. But is there a way to find a minimal example? as you pointed out. $\endgroup$ – Rick Apr 17 at 11:19
  • $\begingroup$ Well, of course...my construction gives an upper bound, so you just have to search every lesser sequence. Not a great method, given how poor the upper bound is. I don't see a quick way to find a minimal solution (though perhaps someone else will). $\endgroup$ – lulu Apr 17 at 11:35
  • $\begingroup$ Thank you again, and really sorry for stating the question poorly. I have edited it, maybe its clearer now? $\endgroup$ – Rick Apr 17 at 11:36

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