SAUDI ARABIAN MATHEMATICAL COMPETITIONS

The answer is YES. We will prove by induction that the procedure can be applied for any positive integer $n$.

The statement is true for $n=1$. Suppose that it is also true for $n=k-1\ge 1$, which means there exists a way to change every number not exceeding $k-1$ from black to white, which we call process $A$.

We shall prove the statement for $k$ with some cases as follows:

1. If $k$ is not a square-free number. Put $k=p_1^{a_1}{p}_2^{a_2}\dots p_v^{a_v}$ and $k^{\prime }=p_1{p}_2\dots p_v$ with $p_1,p_2,\dots ,p_v$ primes. Note that the number of times the color of $k$ and $k^{\prime }$ are changed are equal (since the set of integers not coprime with them are the same), then $k^{\prime }$ can change the color if and only if $k$ can. So after making process $A$ with the sequence $1,2,\dots ,k-1$, the number $k$ will change color from black to white.

2. If $k$ is a square-free number. In case after making process $A$, number $k$ changes color, then we are done. Otherwise, put $k=p_1{p}_2\dots p_t$ and we consider process $B$ to select all divisors other than $1$ of $k$. After that, number $k$ will be affected $2^t-1$ times and will change the color.

Denote $k^{\prime }$ as some divisor greater than $1$ of $k$, and $k^{\prime }$ has $s<t$ prime divisors. Take some number $c$ which is a multiple of $k^{\prime }$ but coprime to $k/k^{\prime }$. After process $A$, number $c$ has color white. And after process $B$, the number of times the color of $c$ is changed is $\left(2^t-1\right)-\left(2^{t-s}-1\right)=2^t-2^{t-s}$ which is an even number. This implies that $c$ will not change color.

So in all cases, we always can find a process to change the color of all numbers from $1$ to $k$ from black to white.

Finally, by replacing $n=10^6$, the problem is solved. $\square$