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> Let $k,M$ be positive integers such that $k−1$ is not squarefree. Prove that there exist a positive real number $α$, such that $⌊α⋅k^n⌋$ and M are coprime for any positive integer $n$. 
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Since $k−1$ is not squarefree, then there exists a prime number $p$ such that $p^2|k−1$. Choose $α=N+\frac{1}{p}$, with $N$ is a positive integer such that $p⋅N+1$ is divisible by all prime factors of $M$ (except $p$ if $p|M$), and $N$ is not divisible by $p$. (we can choose $N$ by using Chinese Remainder Theorem). Then for every positive integer $n$, 
$$⌊α⋅k^n⌋=N⋅k^n+⌊\frac{k^n}{p}⌋=N⋅k^n+\frac{k^n−1}{p}=\frac{k^n⋅(p⋅N+1)−1}{p} $$
Since $p⋅N+1$ is divisible by all prime divisors of $M$, and $⌊α⋅k^n⌋$ is not divisible by $p$, because $N⋅k^n$ is not divisible by $p$ (we consider this in case $p|M$), therefore, $⌊α⋅k^n⌋$ and $M$ are coprime. 


However, if $α$ must be irrational, then I have a feeling that there are no such $α$ that suit the problem's condition. 


So my question is: 



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> Let $p$ be a prime integer, $k$ be a positive integer and $α$ be a positive irrational number. Is it true that there always exists a positive integer $n$ such that $p | ⌊k^n⋅α⌋$ ? 
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Any answers or comments will be appreciated.


(Please let me know if this question should be closed, off-topic or unclear. I may not visit this page frequently, so I may not be able to know what is going on. Sorry for this inconvenience)



Is it true that there always exists a positive integer $n$ such that $p | ⌊k^n⋅α⌋$
I will answer your question (the one in the second orange box).


No, it is not true. For a counterexample, take $p=2$, $k=10$, and
$$\alpha=0.13113311133311113333\dots$$


