1. Determine the greatest number of consecutive five-digit natural numbers, among which there is no palindrome, i.e., a number that reads the same forwards and backwards.
Among 109 consecutive five-digit numbers

$$
10902,10903, \ldots, 10999,11000, \ldots, 11009,11010
$$

there is no palindrome (it is possible to provide other suitable examples of 109 five-digit numbers, we have listed the smallest group of them).

The smallest and largest five-digit palindromes are the numbers 10001 and 99999; there is only one five-digit number before 10001, and there are no such numbers after 99999. We will now show that after each five-digit palindrome \( x, x \neq 99999 \), there follows a five-digit palindrome \( x+100 \) or \( x+110 \) or \( x+11 \). Indeed, if \( x = \overline{a b c b a} \), then in the case \( c \neq 9 \) the number \( x+100 = \overline{a b(c+1) b a} \) is a palindrome, in the case \( c=9 \neq b \) the number \( x+110 = \overline{a(b+1) 0(b+1) a} \) is a palindrome, and finally in the case \( c=b=9 \) (when necessarily \( a \neq 9 \)) the number \( x+11 = \overline{(a+1) 000(a+1)} \) is a palindrome.

Answer. The largest number of sought numbers is 109.

For a complete solution, award 6 points, of which 2 points for a suitable example of 109 numbers and 4 points for the explanation of why in each segment of 110 five-digit numbers there is at least one palindrome. (Most often, this will be the above reasoning that the difference between two adjacent palindromes is either 11, 100, or 110.)