213. For what integer values of $x$ is the number $9 x^{2}+29 x+$ +62 divisible by 16?
213. We immediately notice that for any integer $x$ (whether $x=2t$ or $x=2t+1$), the expression $9x^2 + 29x + 62$ is divisible by 2. Substituting $x=2t$ into it, we find after simplification that $t+1$ is divisible by 2; hence, $t=2t_1-1$, from which $x=4t_1-2$. Substituting $x=2t+1$, we find $t=2t_1$; hence, $x=4t_1+1$. Therefore, $9x^2 + 29x + 62$ is divisible by 4 when $x=4t_1-2$ and $x=4t_1+1$. Substituting $x=4t_1-2$, we find $t_1=2t_2$; hence, $x=8t_2-2$. Substituting $x=4t_1+1$, we find $x=8t_2+5$, which can be written as $x=8t_2-3$. Therefore, the number $9x^2 + 29x + 62$ is divisible by 8 when $x=8t_2-2$ and $x=8t_2-3$. Similarly, we find that $9x^2 + 29x + 62$ is divisible by 16 when $x=16t_3+6$ and $x=16t_3+5$.