1. In this problem, we consider positive integers consisting of 5 digits, where the first and the last digit are not zero. We call such a number a palindrome product if it meets the following two conditions:

- the number is a palindrome (i.e., reads the same from left to right as from right to left);
- the number is a product of two positive integers, one of which reads from left to right the same as the other reads from right to left, such as 4831 and 1384.

For example, 20502 is a palindrome product because $102 \cdot 201 = 20502$ and 20502 is itself a palindrome.

Determine all palindrome products of 5 digits.
1. Note that the two numbers mentioned in the second condition must not end in 0; otherwise, the product would also end in 0. Therefore, the two numbers must have the same number of digits. However, the product of two numbers with four or more digits is too large (it would have at least seven digits), and the product of two numbers with two or fewer digits is too small (it would have at most four digits).

For the second condition, we need to look at products of numbers with three digits, say $\overline{a b c}$ and $\overline{c b a}$. If we work this out, we get: $\overline{a b c} \cdot \overline{c b a}=(100 a+10 b+c) (100 c+10 b+a)=10000 a c+1000 b(a+c)+100\left(a^{2}+b^{2}+\right.$ $\left.c^{2}\right)+10 b(a+c)+a c$. If $a c>9$, then this number has more than 5 digits. Therefore, $a c \leqslant 9$. Suppose $b(a+c)>9$, then the first digit is more than $a c$ and is not equal to the last digit. Therefore, $b(a+c) \leqslant 9$. The same argument applies to

| $a$ | $b$ | $\overline{a b c} \cdot \overline{c b a}$ |  |
| :---: | :---: | :---: | :---: |
| 1 | 0 | 1 | 10201 |
| 2 | 0 | 2 | 20502 |
| 2 | 0 | 1 | same |
| 1 | 1 | 40804 |  |
| 1 | 2 | 23632 |  |
| 2 | 1 | 1 | same |
| 2 | 1 | 2 | 44944 |
| 1 | 1 | 14641 |  |
| 1 | 2 | 26962 |  |
| 2 | 2 | 1 | same |

$a^{2}+b^{2}+c^{2}$. So we know that $1+b^{2}+1 \leqslant a^{2}+b^{2}+c^{2} \leqslant 9$ thus $b=0, b=1$ or $b=2$.

For $b=0$, we must have $a^{2}+c^{2} \leqslant 9$, so (since $a, c \geqslant 1$) $1 \leqslant a \leqslant 2$ and similarly for $c$.

For $b=1$, the conditions become $a+c \leqslant 9$ and $a^{2}+c^{2} \leqslant 8$, while also $1 \leqslant a c \leqslant 9$. The second condition gives $1 \leqslant a \leqslant 2$ and similarly for $c$.

For $b=2$, the conditions become $a+c \leqslant 4$ and $a^{2}+c^{2} \leqslant 5$. So there are three possibilities for $(a, b)$.

Now we will go through all the possibilities in the table above. It turns out that there are a total of eight different palindromic products: 10201, 12321, 14641, 20502, 23632, 26962, 40804, and 44944.