If $m$ is divisible by $10$, then $k$ would have to end with a $0$, so it is impossible. Otherwise, take $k=(m(10^N+1))^2$ with $N$ big enough (greater than $2021$ plus the number of digits of $m^2$). Therefore the only zero taker numbers are the positive integers not divisible by 10

Mattysal wrote: A positive integer $m$ is said to be $\emph{zero taker}$ if there exists a positive integer $k$ such that: $k$ is a perfect square; $m$ divides $k$; the decimal expression of $k$ contains at least $2021$ '0' digits, but the last digit of $k$ is not equal to $0$. Find all positive integers that are zero takers.

My construction

Existence

Final answer