Problem 1119

Solution: Zero is a red number: indeed, if 0 were white, given that there exists a red number $x$, we would have that $0+x=x$ is white by the second property, contradiction.

One is a white number: indeed, if one were red, given that there exists a white number $y$, we would have that $y \cdots 1=y$ is red by the third property, contradiction.

If there are no red numbers other than zero, the thesis is trivial. Otherwise, let $k$ be the smallest red number greater than zero. Then every number that is not a multiple of $k$ is white: indeed, if $n$ is not a multiple of $k, n$ can be written in the form $n=q k+r$ with $0<r<k$. We use induction on $q$. If $q=0, n$ is white by assumption. Assuming the hypothesis is true for $q-1$, we have $n=[(q-1) k+r]+k$, which is white by the second property.

By the third property, every multiple of $k$ of the form $j \cdot k$, with $j$ not divisible by $k$, is red. Suppose now that $n$ is a multiple of $k$ of the form $j \cdot k$ with $j=l k$ divisible by $k$, i.e., that $n$ is of the form $l \cdot k^{2}$. From the equality $k+l \cdot k^{2}=(1+l k) \cdot k$ we have, by the second property, that in this case $n$ must also be red. Therefore, the red numbers are all and only the multiples of $k$. In this case, both the assumptions of the problem and the thesis are trivially satisfied.