211. For what integer values of $x$ is the number $x^{2}+2 x+6$ divisible by 15?
211. First, let's find the integer values of $x$ for which $x^{2}+2 x+6$ is divisible by 3, which is equivalent to the divisibility of $x^{2}+2 x$ by 3; we find $x=3 t$ and $x=3 t+1$. From these values of $x$, we take those for which $x^{2}+2 x+6$ or $x^{2}+$
$+2 x+1$ is divisible by 5. Substituting $x=3 t$ into the latter, we get that $(3 t+1)^{2}$, and therefore $3 t+1$ is divisible by 5; hence $t=5 t_{1}-2$, so $x=15 t_{1}-6$.

Substituting $x=3 t+1$ into the expression $x^{2}+2 x+1$, we get that $(3 t+2)^{2}$, and therefore $3 t+2$ is divisible by 5; hence $t=5 t_{1}+1$, so $x=15 t_{1}+4$. Thus, $x^{2}+2 x+6$ is divisible by 15 when $x=15 t-6$ and $x=15 t+4$.