Is there such an integer? $\{a_n\}_{n=1}^{\infty}$ and $\{b_n\}_{n=1}^{\infty}$ are two infinite sequences of natural numbers such that for all $n$, $0\leq b_n<a_n$. Is it possible to exist an integer $k$ such that for all $n$, $k\equiv b_n(\mbox{mod } a_n)$?
Thanks for your answer!

Not always possible. To see this, let $k$ be an integer. Then $k$ has the property iff
for every $n \geq 1$ there is an integer $q$ such that
$$k = qa_{n} + b_{n}.$$
If the sequence $(b_{n})$ is not such that $b_{n} =$ const. for all $n \geq 1$ and if $a_{n} = 0$ for all $n \geq 1,$ then there is no $k$ such that $k$ has the property. 
