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
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.