There are 900 three-digit * * *, ranging from 100 to 999. Arrange these 900 numbers from small to large and divide them into 90 groups. For any group, let the sum of the three digits of the first number be A, then the following are A+ 1, A+2, ... But only two of the continuous natural numbers of 10 can be divisible by 5, so there must be and only two of 10 meet the conditions, so there are 2× 90 =/kl.

For example: 3 10, 3 1 1, 3 12, ... 3 19, the sum of the numbers in this group 10 is 4, 5, 6, .../kloc-.

2. Look at the following rules first.

1 1? = 12 1, and the sum of its digits is 1+2+ 1 = 4.

1 1 1? = 1232 1, and the sum of its digits is 1+2+3+2+ 1 = 9.

1 1 1 1? = 123432 1, and the sum of its digits is1+2+3+3+2+1=16.

So the sum of digits of the square of 1 1 ... 1 (1989 1) is1+2+3+4+...+1989+ 988+ 1989= 1989× 1989= 1989?