Analysis and solutions:

-It is known that the four-digit 3AA 1 is divisible by 3, so its sum of numbers (3+A+A+ 1) must be a multiple of 3. Because a is a number, it can only be 0, 1, 2, 3, ..., 9, and the maximum value can only be 9. If A=9, then 3+A+A+ 1 = 22, 22 < 27, then the sum of digits of 3AA 1 can only be a multiple of 1 or twice of 3, that is, 3 or 6.

When 3+A+A+ 1 = 3, A=0, which is irrelevant.

When 3+A+A+ 1 = 6, A= 1, which is in line with the question, so A stands for 1, and this four-digit number is 3 1 1.