5. We have cards with numbers 5, 6, 7, ..., 55 (each card has one number). What is the maximum number of cards we can select so that the sum of the numbers on any two selected cards is not a palindrome? (A palindrome is a number that reads the same backward as forward.)
SOLUTION. To be able to express ourselves more concisely, we will choose numbers directly, not cards.

First, note that for the sum $s$ of any two given numbers, $11=$ $=5+6 \leqq s \leqq 55+54=109$. Among the numbers from 11 to 109, the palindromes are precisely all multiples of 11 and the number 101. Now, realize that the divisibility of the sum of two numbers by a given number $d$ (in our case, $d=11$) depends only on the remainders of the two addends when divided by the given $d$. This useful rule will be applied by dividing all the given numbers from 5 to 55 into groups according to their remainders when divided by 11, and writing these groups into rows so that the sum of two numbers from different groups on the same row is divisible by 11; we will explain the meaning of the parentheses at the end of each row shortly.

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At the end of each row, we have added the maximum number of numbers that can be selected from that row without the sum of any two of them being a multiple of 11. For example, in the third row, we have a quintet of numbers with a remainder of 8 and a quartet of numbers with a remainder of 3. It is clear that we cannot simultaneously select one number from each of these groups (their sum would be a multiple of 11), but we can simultaneously select all five numbers from the quintet (the sum of any two of them will give the same remainder as the sum $8+8$, i.e., a remainder of 5). Let us add that the scheme of six rows has another huge advantage for us: the sum of any two numbers from different rows is not a multiple of 11 (neither is the sum of their two remainders).

From the given division of all the given numbers into six rows, it follows that we cannot select more than $5 \cdot 5+1=26$ numbers in the required way. However, if we selected 26 numbers, there would have to be one of the numbers 49 or $50 \mathrm{a}$ and from the other four rows, one number each from 51, 52, 53, and 54 - then we would get the palindrome $49+52$ or $50+51$. Therefore, it is not possible to select more than 25 numbers, and a selection of 25 numbers is possible: on the first five rows, we can select, for example, all the numbers from the left groups except for the number 52 and one number (e.g., 11) from the last row. Then the sum of any two selected numbers will not be divisible by 11 (thanks to the grouping of numbers), let alone equal to the last "critical" number, the palindrome 101 (which is why we excluded 52 when choosing 49).

Answer. The largest number of cards that can be selected in the required way is 25.