Given the equation $x^{2}+(a-6) x+a=0$ $(a \neq 0)$ has two integer roots, find the integer value of $a$.

---

The equation $x^{2}+(a-6) x+a=0$ $(a \neq 0)$ is known to have two integer roots. Try to find the integer value of $a$.
Let the two integer roots of the equation be \(x_{1}, x_{2}\), and \(x_{1} \geqslant x_{2}\). By Vieta's formulas, we have:
\[
\left\{\begin{array}{l}
x_{1}+x_{2}=6-a, \\
x_{1} x_{2}=a .
\end{array}\right.
\]

(1)
\[
\begin{array}{l}
+ \text { (2): } x_{1} x_{2}+x_{1}+x_{2}=6, \\
\therefore \quad\left(x_{1}+1\right)\left(x_{2}+1\right)=7 .
\end{array}
\]

From this, solve for \(x_{1}, x_{2}\), and then combine with the problem statement to get \(a=16\).