## Discrete Mathematics with Applications 4th Edition

Yes, $3$ divides $(3k+1)(3k+2)(3k+3)$. To see this, note that $3k+3=3(k+1)$ and that $3|3(k+1)$, since $3(k+1)=3\times(k+1)$, where $k+1$ is an integer because $k$ and $1$ are both integers. By Theorem 4.3.3 (transitivity of divisibility), since $3|(3k+3)$ and $(3k+3)|(3k+1)(3k+2)(3k+3)$, it must be that $3|(3k+1)(3k+2)(3k+3)$.
The statement that $3$ divided $3k+1)(3k+2)(3k+3)$ is written symbolically as $3|(3k+1)(3k+2)(3k+3)$. By definition, for two integers $a$ and $b$ where $a\ne0$, $a|b$ if and only if $b=na$ for some integer $n$. Here, $a=3$, $b=(3k+1)(3k+2)(3k+3)$, and $n=(k+1)^{2}(3k+2)$. Theorem 4.3.3 states that $a|b$ and $b|c$ imply $a|c$.