3. Find the number of four-digit numbers in which all digits are different, the first digit is divisible by 2, and the sum of the first and last digits is divisible by 3.
Solution. The first digit can be 2, 4, 6, or 8. If the first is 2, the last can be 1, 4, 7; if the first is 4, the last can be 2, 5, 8; if the first is 6, the last can be 0, 3, (6 does not fit), 9; if the first is 8, the last can be 1, 4, 7. In total, there are 3 + 3 + 3 + 3 = 12 options for the first and last digits. For each of these options, there are 8 * 7 ways to choose the two middle digits. In total, 56 * 12 = 672 ways.

## Criteria.

Errors. Numbers of the form 6 ** 6 are incorrectly counted - two points are deducted.

Without explanation, it is stated that each first digit corresponds to three variants of the last - one point is deducted (this is not obvious, as it is only due to the impossibility of 6).

Numbers starting with zero are incorrectly counted - two points are deducted.

Arithmetic errors in the correct solution method - one point is deducted for each error.

Advancements. If it is written that the number of pairs (second digit, third digit) is not 7 * 8, but 9 * 9 or 8 * 8, etc., the problem is considered unsolved and no more than 2 points are given.