1. Determine the number of all five-digit palindromes that are divisible by 37. (A palindrome is a number that reads the same backward as forward in decimal notation.)
1. Every five-digit palindrome $p$ can be written in the form $p=\overline{a b c b a}$, where $a, b, c$ are digits in the decimal system, $a \neq 0$. From the expression

$$
p=10001 a+1010 b+100 c=37(270 a+27 b+3 c)+11(a+b-c)
$$

it follows that $p$ is divisible by 37 if and only if the number $a+b-c$ is divisible by 37. Given that $a, b, c$ are digits $(a \neq 0)$, we have $-8 \leqq a+b-c \leqq 18$. Therefore, the number $a+b-c$ is divisible by 37 if and only if $a+b-c=0$, or $c=a+b$. The digits $a, b$ must therefore satisfy the condition $a+b \leqq 9$.

For each $a \in\{1,2,3, \ldots, 9\}$, the digit $b$ can be chosen in $10-a$ ways such that $a+b \leqq 9$ (i.e., $b \in\{0,1,2, \ldots, 9-a\}$). The digit $c$ is then uniquely determined as the sum $a+b$. Therefore, there are 9 palindromes with the digit $a=1$, 8 palindromes with the digit $a=2$, and so on; finally, for the digit $a=9$, there is exactly one palindrome.

The total number of five-digit palindromes that are divisible by 37 is thus

$$
9+8+7+\ldots+1=45
$$

Full solution is worth 6 points, finding the condition $c=a+b$ is worth 4 points.