Problem #5

Question: Prove that for any integer $n$, at least one of the integers $n$, $n+2$, $n+4$ is divisible by 3.

Claim: for any integer $n$, at least one of the integers $n$, $n+2$, $n+4$ is divisible by 3.
Proof: by the General Division Theorem, every integer number $n$ can be presented as $3q+r$, where $0 \leq r < 3$, so $n$ can be in one of those forms:
$$\begin{equation} 3q+0 \end{equation}$$
$$\begin{equation} 3q+1 \end{equation}$$
$$\begin{equation} 3q+2 \end{equation}$$
Suppose we get an arbitrary $n$ in the form $3q+0$. It's clear that in this case $n$ is divisible by 3, by the divisibility property.
Suppose we get an arbitrary $n$ in the form $3q+1$. $n$ is not divisible by 3. But $n+2=(3q+1)+2=3q+3=3(q+1)$ is, by divisibility property.
Suppose we get an arbitrary $n$ in the form $3q+2$. $n$ is not divisible by 3. $n+2=(3q+2)+2=3q+4$ is not divisible by 3. But $n+4=(3q+2)+4=3q+6=3(q+2)$ is, by divisibility property.
We have got an integer $n$ in all forms it can be and proved that at least one of the integers $n$, $n+2$, $n+4$ is divisible by 3, thus for any integer $n$ the initial claim has been proved. $\blacksquare$