There is a property that two adjacent natural numbers are prime numbers. Such as 49 and 50.
Another property is that two adjacent odd numbers are prime numbers. Such as 49 and 5 1.
If A wants to win, just make sure that the last two numbers are adjacent.
There are 1989 numbers in total.
So A only needs to erase an even number first.
Then, then see how b wipes it,
If the number erased by B is greater than the first number of A, then B is even, then A is less than B, and if B is odd, then A is greater than B. ..
If b erases a number smaller than the first number of a, then b erases an odd number, then a erases a number smaller than this number, and if b erases an even number, a erases a number larger than this number.
Rules, erased numbers will flow out of a gap, so it is to keep the number of numbers in the gap even.
Of course, there may be other ways, but this is the way I can think of, and A can win. Formula, no.
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For example: 2, 3, 4, 5, 6, 7, 8, 9, 10,1,12.
First erase an even number, such as 6.
1. Then look at B. If B erases 8, because 8 is bigger than 6 and even. So "A will then erase this number -8, which is smaller than B", which is 7.
2. Then B erases 9, because 9 is bigger than 6, which is an odd number. So "A will then erase this number -9, which is one number larger than B", that is, 10.
3. Then B erases 5, because 5 is smaller than 6 and is an odd number. So "A will then erase this number -5, which is smaller than B", which is 4.
4. Then B erases 2, because 2 is smaller than 6 and is even. So "A will then erase this number -2, which is a number larger than B", which is 3.
In this way, 1 1 and 12 are left after the nail polish, which are prime numbers, so A wins. If it is extended to 2, 3, 4, 5... 1990, the properties are the same. Also in accordance with these four principles.