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.
Problem
Source: ITAMO 2021 - Problem 1
Tags: number theory
cadaeibf
08.05.2021 14:02
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
Sprites
27.07.2021 18:38
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.
So similar to @above,We will show this for $m(10^N+1)^2$
We will show that $m(10^N+1)^2$ has atleast $2021$ $0's$ for some $N$ which is extremely large.
So here $$m(10^N+1)^2=m*10^N+2*m*10^N+1$$will have atleast 2021 0's for sufficiently large$N$
It works for any $m$ until and unless $10|m$