1. Let $x$ be a positive integer, and $x<50$. Then the number of $x$ such that $x^{3}+11$ is divisible by 12 is $\qquad$.
$$
\begin{array}{l}
x^{3}+11=(x-1)\left(x^{2}+x+1\right)+12 \\
=(x-1) x(x+1)+(x-1)+12 .
\end{array}
$$

Notice that $6 \mid(x-1) x(x+1)$, so $6 \mid(x-1)$;
also, $2 \mid \left(x^{2}+x+1\right)$, hence $12 \mid (x-1)$.
Since $x<50$, we have
$$
x-1=0,12,24,36,48 \text {. }
$$

Thus, $x=1,13,25,37,49$, a total of 5.