5. From the set $\{1,2, \cdots, 105\}$, randomly select an element $a$, such that the equation $x^{2}+a x+6 a=0$ has only integer solutions. The probability is $\qquad$.
5. $\frac{1}{21}$.

Let the integer solutions of the equation $x^{2}+a x+6 a=0$ be $m, n$ $(m \leqslant n)$. By Vieta's formulas, we have
$$
\begin{array}{l}
m+n=-a, m n=6 a \\
\Rightarrow m n=-6(m+n) \\
\Rightarrow(m+6)(n+6)=36 .
\end{array}
$$

Since $36=1 \times 36=2 \times 18$
$$
=3 \times 12=4 \times 9=6 \times 6 \text {, }
$$

the solutions $(m, n)$ satisfying $m \leqslant n$ are
$$
\begin{array}{l}
(-42,-7) 、(-24,-8) 、(-18,-9) 、 \\
(-15,-10) 、(-12,-12) 、(-5,30) 、 \\
(-4,12) 、(-3,6) 、(-2,3) 、(0,0),
\end{array}
$$

and the corresponding values of $a=-(m+n)$ are $49, 32, 27, 25, 24, -25, -8, -3, -1, 0$.
Thus, there are 5 values of $a$ that satisfy $a \in\{1,2, \cdots, 105\}$.
Therefore, the required probability is $\frac{5}{105}=\frac{1}{21}$.