18. (3 points) If $a b c+c b a=1069$, then there are $\qquad$ such $\underline{a b c}$.

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
18. (3 points) If $\underline{a b c}+\underline{c b a}=1069$, then there are 8 such $\underline{a b c}$.
【Analysis】Solution: (1) $a b c+c b a=100 a+10 b+c+100 c+10 b+a=101 a+101 c+20 b=101 \times(a+c)+20 b=1069$;
(2) According to the unit digit 9 of $101 \times(a+c)+20 b=1069$, we know: the unit digit of $a+c$ must be 9; since the maximum value of $a$ and $c$ does not exceed 9, therefore $a+c=9$, which means $b$ must be 8;
(3) Analyze how many cases there are for $a+c=9$: Since $a$ and $c$ both appear in the first position, they cannot be 0. Then, when $a=1, 2, 3, 4, 5, 6, 7, 8$, the values of $c$ can be derived, solving the problem.
【Solution】Solution: (1) From $\underline{a b c}+\underline{c b a}=1069$, we get $a b c+c b a=100 a+10 b+c+100 c+10 b+a=101 a+101 c+20 b=101 \times(a+c)+20 b=1069$;
(2) Since the unit digit of $101 \times(a+c)+20 b=1069$ is 9, the unit digit of $a+c$ must be 9; since the maximum value of $a$ and $c$ does not exceed 9, therefore $a+c=9$, which means $b$ must be 8;
(3) Since $a+c=9$, and since $a$ and $c$ both appear in the first position, they cannot be 0. When $a=1$, $c=8$; when $a=2$, $c=7$; when $a=3$, $c=6$; when $a=4$, $c=5$; when $a=5$, $c=4$; when $a=6$, $c=3$; when $a=7$, $c=2$; when $a=8$, $c=1$;
Therefore, each combination corresponds to one case, so there are 8 different numbers for $a b c$.
In summary, there are 8 such $a b c$.
The answer is: 8.
【Comment】This problem starts from the relationship $101 \times(a+c)+20 b=1069$, and the unit digit 9 of $101 \times(a+c)+20 b=1069$ leads to the conclusion that the unit digit of $a+c$ must be 9, which is the key to solving the problem.