If there is a 1994 digit a divisible by 9, the sum of its digits is a, the sum of its digits is B, and the sum of its digits is C, then C =?

If there is a 1994 digit a divisible by 9, the sum of its digits is a, the sum of its digits is B, and the sum of its digits is C, then C =?

A number divisible by 9, the sum of the numbers divisible by 9
So 1994 digit a can be divided by 9,
A the biggest is 999 The minimum is 1000 8 [1992 0]
The number of a and a must be a multiple of 9, the maximum is 1994 * 9 = 17946, the minimum is 9
For a, the sum of his numbers can also be divided by 9
A may be an integer divided by 9 between 9 and 17946
It is easy to know that the sum of the digits B of a, the maximum is 9 * 4 = 36 [when a = 9999], and the minimum is 9
Similarly, for B, the sum of his numbers can be divided by 9
B may be a number between 9 and 36 divided by 9, that is, B = 9, 18, 27, 36
Therefore, the sum of the numbers of B is equal to and can only be equal to 9