Let's denote a natural number $n$ as interesting if both $n$ and $n+2023$ are palindromes, that is, numbers that read the same from left to right and from right to left. Find the smallest and the largest interesting numbers.
Let's find the smallest interesting number. First, let's see if an interesting \( n \) can be so small that \( n + 2023 \leq 2999 \), i.e., \( n \leq 976 \). In this case, \( n + 2023 \) is between 2023 and 2999, and thus starts with 2 - and therefore, it also ends with 2. Since the sum \( n + 2023 \) ends with 2, \( n \) itself ends with 9. Note that \( n \) cannot be a single-digit number (since \( 2023 + 9 = 2032 \) is not a palindrome), nor can it be a two-digit number (then it would be 99, and \( 2023 + 99 = 2122 \) is not a palindrome). Therefore, \( n \) is a three-digit number, and has the form 9?9. But then the sum \( n + 2023 \) is at least 2900, and thus must equal 2992. Subtracting, we find \( n = 2992 - 2023 = 969 \). Thus, \( n = 969 \) is the smallest interesting number.

Now let's find the largest interesting number. Suppose \( n \) is a five-digit number or more. When adding 2023, either the number of digits increases, and then the first digit changes from 9 to 1 (with a carry-over), or the number of digits remains the same, and then the first digit either does not change or changes by 1. But the last digit changes by 3 (possibly with a carry-over) - a contradiction.

For \( 2023 + abba \) to be a palindrome, it must be a five-digit number. Otherwise, in the hundreds place, the digit \( b \) increases by 1 or remains the same, while in the tens place, it increases by 2 or 3. Therefore, the first digit of the number \( 2023 + abba \) is 1, and the last digit will be 1 only if \( a = 8 \). In the thousands place of the number \( 2023 + 8bb8 \), there can be either 0 or 1. To get the same digit in the tens place, \( b = 8 \) or \( b = 7 \). The number 8888 does not work: \( 8888 + 2023 = 10911 \), but 8778 is interesting: \( 8778 + 2023 = 10801 \).

## Answer

969 and 8778.

## [ Inequalities. Case Analysis ] Problem 35063 Topics: [ Ordering by Increasing (Decreasing) $]$

Given 10 natural numbers, not exceeding 91. Prove that the ratio of some two of these numbers belongs to the interval \([2/3, 3/2]\).

## Hint

Order the numbers in increasing order. Can each number exceed the previous one by more than 1.5 times?

## Solution

Suppose for some numbers \( a_{1} \leq a_{2} \leq \ldots \leq a_{10} \leq 91 \) the statement is not true. Then each of the numbers (except the first) is more than 1.5 times the previous one.

\( a_{1} \geq 1 \) (the numbers are natural). From this, we sequentially obtain: \( a_{2} > 3/2 \), i.e., \( a_{2} \geq 2 \); \( a_{3} > 3/2 \cdot 2 = 3 \), i.e., \( a_{3} \geq 4 \); \( a_{4} > 6 \), i.e., \( a_{4} \geq 7 \);

\( a_{5} \geq 11 \); \( a_{6} \geq 17 \); \( a_{7} \geq 26 \); \( a_{8} \geq 40 \); \( a_{9} \geq 61 \); \( a_{10} \geq 92 \). Contradiction.