1.49 Find the values of $a$ such that the roots of the equation $x^{2}-a x+9 a=0$ are integers.
(Recommended by the Soviet Ministry of Education, 1990)
［Solution］Let $x_{1}, x_{2}$ be the integer roots of the equation
$$
x^{2}-a x+9 a=0
$$

Then, by Vieta's formulas, we have
$$
\begin{array}{c}
x_{1}+x_{2}=a, \\
x_{1} x_{2}=9 a .
\end{array}
$$

Thus,
$$
\begin{array}{c}
x_{1} x_{2}=9\left(x_{1}+x_{2}\right) . \\
x_{1} x_{2}-9 x_{1}-9 x_{2}+81=81, \\
\left(x_{1}-9\right)\left(x_{2}-9\right)=81 .
\end{array}
$$

Notice that, 81 has the following 6 factorizations:
$$
\begin{aligned}
81 & =1 \cdot 81=(-1) \cdot(-81)=3 \cdot 27 \\
& =(-3) \cdot(-27)=9 \cdot 9=(-9) \cdot(-9) .
\end{aligned}
$$

Therefore,
$$
a=x_{1}+x_{2}=\left(x_{1}-9\right)+\left(x_{2}-9\right)+18
$$

We can obtain the following 6 numbers:
$$
\pm 82+18, \quad \pm 30+18, \quad \pm 18+18 .
$$

That is, $a=100,-64,48,-12,36,0$.
Thus, when $a$ takes these 6 integer values, the given equation has integer roots.