Determine all positive integers $M$ for which the sequence $a_{0}, a_{1}, a_{2}, \ldots$, defined by $a_{0}=\frac{2 M+1}{2}$ and $a_{k+1}=a_{k}\left\lfloor a_{k}\right\rfloor$ for $k=0,1,2, \ldots$, contains at least one integer term. (Luxembourg) Answer. All integers $M \geqslant 2$.

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Determine all positive integers $M$ for which the sequence $a_{0}, a_{1}, a_{2}, \ldots$, defined by $a_{0}=\frac{2 M+1}{2}$ and $a_{k+1}=a_{k}\left\lfloor a_{k}\right\rfloor$ for $k=0,1,2, \ldots$, contains at least one integer term. (Luxembourg) Answer. All integers $M \geqslant 2$.
Define $b_{k}=2 a_{k}$ for all $k \geqslant 0$. Then 
$$ b_{k+1}=2 a_{k+1}=2 a_{k}\left\lfloor a_{k}\right\rfloor=b_{k}\left\lfloor\frac{b_{k}}{2}\right\rfloor . $$ 
Since $b_{0}$ is an integer, it follows that $b_{k}$ is an integer for all $k \geqslant 0$. Suppose that the sequence $a_{0}, a_{1}, a_{2}, \ldots$ does not contain any integer term. Then $b_{k}$ must be an odd integer for all $k \geqslant 0$, so that 
$$ b_{k+1}=b_{k}\left\lfloor\frac{b_{k}}{2}\right\rfloor=\frac{b_{k}\left(b_{k}-1\right)}{2} $$ 
Hence 
$$ b_{k+1}-3=\frac{b_{k}\left(b_{k}-1\right)}{2}-3=\frac{\left(b_{k}-3\right)\left(b_{k}+2\right)}{2} $$ 
for all $k \geqslant 0$. Suppose that $b_{0}-3>0$. Then equation (2) yields $b_{k}-3>0$ for all $k \geqslant 0$. For each $k \geqslant 0$, define $c_{k}$ to be the highest power of 2 that divides $b_{k}-3$. Since $b_{k}-3$ is even for all $k \geqslant 0$, the number $c_{k}$ is positive for every $k \geqslant 0$. 
Note that $b_{k}+2$ is an odd integer. Therefore, from equation (2), we have that $c_{k+1}=c_{k}-1$. Thus, the sequence $c_{0}, c_{1}, c_{2}, \ldots$ of positive integers is strictly decreasing, a contradiction. So, $b_{0}-3 \leqslant 0$, which implies $M=1$. 
For $M=1$, we can check that the sequence is constant with $a_{k}=\frac{3}{2}$ for all $k \geqslant 0$. Therefore, the answer is $M \geqslant 2$.