6. The number 1 is written on the board. If the number a is written on the board, it can be replaced by any number of the form a+d, where $d$ is coprime with a and $10 \leq d \leq 20$. Is it possible to obtain the number $18!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot 18$ on the board after several such operations? (I. Rubanov)
First solution. Note that the number 18!-19 ends with 1. We will add 10 to the number on the board. Each time, we will get a number ending in 1, and thus coprime with the number 10, so the operation is possible. Eventually, the number 18!-19 will appear on the board. We will add 19 to it and get 18!.

Second solution. It is clear that 18! is not divisible by 19. Then 18!-19k is not divisible by 19 for any natural k. Now, if we learn to obtain numbers that have all possible non-zero remainders when divided by 19, then by adding 19 to one of them, we will be able to get 18!. (In particular, it would be enough to learn to obtain some 19 consecutive numbers.) We will learn how to do this. The numbers from 11 to 21 are obtained in one operation. Numbers of the form 22+n for 0<n<10 are obtained as 1+10+(11+n). The number 22 cannot be obtained, but the number 41 of the same remainder can, for example, 41=1+10+16+14.

Third solution. By Wilson's theorem, 18! ≡ 18 (mod 19). Therefore, it is enough to get 18=1+17 on the first step, and then add 19 each time.

Fourth solution. 18!-19 ≡ 17 (mod 18). Therefore, it is enough to get 17=1+16 on the first step and then add 18 until we get 18!-19. Now we can add 19.

Criteria. It is proven that 18! can only be obtained from 18!-19, and there is no further substantial progress - 1 point.

A completely correct algorithm without any justification - 4 points. The problem is reduced to obtaining the remainder of 18! when divided by 19, but the remainder is found incorrectly - no more than 1 point.

To justify the algorithm from the first solution, it is necessary to prove the following three (or equivalent) statements: (1) all intermediate numbers are coprime with 10; (2) the last number is coprime with 19; (3) 18! ends in 0 (a reference to this statement is sufficient at the right moment). For the absence of each of the justifications in (1) and (2) and the absence of a reference in (3), 1 point is deducted.

For the second solution. The justification that 18!-19k is coprime with 19 is missing - minus 1 point. The problem is reduced (with justification) to obtaining numbers that give all non-zero remainders when divided by 19, but no more than three remainders are not obtained or are obtained incorrectly - 4 points. The same, more than three remainders are not obtained or are obtained incorrectly - 3 points.