8.5. The first two digits of a natural four-digit number are either each less than 5, or each greater than 5. The same can be said about the last two digits. How many such numbers are there? Justify your answer.
Solution: There are $4^{2}=16$ two-digit numbers where both digits are greater than 5, and there are $4 \cdot 5=20$ (the first digit is not zero) where both digits are less than 5. In total, there are $16+20=36$ such numbers. This is the conclusion about the first two digits of the number. The last two digits will provide the same number of options plus another 5: 00, 01, 02, 03, and 04, totaling 41 options. The total number of numbers is $36 \cdot 41=1476$.

Note: 1476 numbers.

Recommendations for checking:

| included in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| Answer is incorrect due to arithmetic errors | 6 points |
| Answer is fully justified but not written as a natural number | no points deducted |
| In a correct reasoning, it is not taken into account that the first digit of the number is never zero | 5 points |
| In a correct solution approach, the number of pairs of the first (and/or last) two digits of the number is incorrectly calculated; the error is not in arithmetic | 4 points |
| Attempt to separately count the number of pairs of the first two digits of the number and the last two digits of the number without further progress in the solution | 2 points |
| Correct answer without justification or by enumeration, if the completeness of the latter is not proven and not obvious | 1 point |
| Attempts to solve by enumeration, not leading to the correct answer | 0 points |