{even, even, even}-{even, even, odd}-{even, even, odd}- ...

If you remove an even number and leave an even number, subtract the sum of the other two numbers with 1 to get an even number, so the two even numbers are 1 odd;

If you remove an odd number and leave two pairs, subtract the sum of the other two numbers with 1 to get an odd number, then the two pairs are 1 odd.

So no matter how to take it, it must be an even, even and odd combination. (17,75,91) is an odd combination, so it is impossible.

3. First of all, from the above two questions, we can know that the process of this rule has the following two possibilities:

{even, even, even}-{even, even, odd}-{even, even, odd}- ..............

{odd, odd, odd}-{odd, odd, odd}-{odd, odd, odd}- ...

Therefore, to get the number 20 10, a must be an even number. ....................................................................................................................................................

{a, a, a}-{a, a, 2a- 1}-{a, 3a-2, 2a- 1} or {a, a, 2a- 1}-

It is found that except the numbers given at the beginning, all other numbers appear in the form of xa-(x- 1), such as 2A- 1, 3A-2, 5A-4, ...

Let xa-(x- 1)=20 10, then (x- 1)a=2009.

From above, A is odd, but from the formula 1, A is even, which is contradictory. So if a is not equal to 20 10, 20 10 will not be generated later.

That is to say, except that 20 10 can be generated at the beginning of a=20 10, it is impossible to generate 20 10 when a is other even numbers.

It must be right. If you don't understand, call me. I really admire myself.