A: Good representative of mathematics 109. If all four digits are ABCD, you can get ABCD-CDC=ABC. Regardless of whether CDC and ABC are arbitrary positive numbers, the sum of two three-digit numbers cannot exceed 2000. Therefore, the topic A= 1 can be solved 1bc = 1bc. Looking at the percentile of the formula again, B is definitely less than C. If B is greater than C, then 1BCD-CDC is four digits, which is inconsistent with the meaning of the question. And because the percentile of the remainder is 1 percentile B- percentile C= 1, and b is less than c, so b can only take 0, then C=9 is D-C=C, and C=9, then it can be concluded that D=8 A= 1, and B = 0. C = 9。