Recursive Digit Sums

Since, $A$ is a multiple of $10!$ , $A$ is a multiple of 9.

So, the digit sum of $A$ must be divisible by 9. And,so $B$ is divisible by 9.
So, the digit sum of $B$ must also be divisible by . So, $C$ is divisible by 9.
So, the digit sum $C$ is also divisible by 9. So, $D$ is divisible by 9.

$A$ is a 2001 digit number that is divisible by 9. We will now prove that no matter how large a 2001 digit number $A$ is, if it is divisible by 9, $D$ will always be 9.

The maximum digit sum for a 2011 digit number is $9\times 2001 = 18009$ .
So, the maximum number of digits of $B$ cannot be greater than 5 and $B\leq 18009$ . Though, $B$ not greater than 18009, it doesn't necessarily mean that the digit sum of $B$ won't be greater than the digit sum of 18009 . But, since the maximum digits of $B$ is 5, it is safe to say that the maximum value of $C$ is at most $9 \times 5 = 45$ .

So, maximum number of digits that $C$ might have is 2. Now, $C$ is divisible by 9 and $C\leq 9$ . All the multiples of 9 in the interval $[9,45]$ have a digit sum of 9. So, the digit sum of $C$ will always be 9.

So, $D$ can be nothing but 9.

[Note: There are ways to improve this bounds on $B$ and $C$ slightly (as Adrian points out in his comment), but it is not necessary.
Pop Quiz: What is the largest number, in place of $2001$ , that we could use in the question, for the answer to still always be 9? - Calvin]

10!=3628800. Multiplying 10! by 10^1994 gives a 2001 digit number with 36288000.... The only nonzero digits are 36288. 3+6+2+8+8=27. 2+7=9, and 9=9, which is our desired answer.

Since $A$ is a multiple of $9$ , we have that $B, C$ and $D$ are a multiple of 9. Thus $B \leq 9\times 2001 = 18009$ which implies that $C \leq 1 + 9 + 9 + 9 + 9 = 37$ . From this we get $D \leq 3 + 9 = 12$ . Since $D$ is divisible by 9 and the digit sum is greater than 0, we have that $D = 9$ . Hence the unit's digit of $D$ is 9.

Given in the problem number A, which has 2001 digits, is divisible by 10! The number of digits A has does not matter since we are only concerned about what the number A is a multiple of.

Since number A is multiple of 10! We do some quick mental math and find 10!=3628800 and 10!x2=7257600. And since when you add the digits of 10! it is equal to the sum of the digits of 10!x2 , hence number A digit's sum must have the same sum as 10!

Therefore 3+6+2+8+8=27 from 10! gives us B, and 2+7=9 gives us C, and 9+0=9 which is D, hence 9 is the answer.

10!=362880 then A being a 2001 digit no. can be 3.6288*10^21 in this case , B=3+6+2+8+8=27 C=2+7=9 D=9 Therefore unit's place digit of D=9

10!= 3628800 , and its a 7 digit number.We make it 2001 digit number by multiplying it by 10^1994 , and it digit sum stays the same 27 (3+6+2+8+8+0+0...+0). C=digit sum of 27 (2+7) which is 9. D=digit sum of 9 which is 9. Units of a singular number equals to it self , and that's the end of task.

As 10! = 3 628 800,we let A=3628800000...000,and B=27,D=C=9

Since it is not mentioned about any other property of the number apart from the fact that it is a multiple of 10! . We can actually infer that any number with 2001 digits that is a multiple of 10! , such that we follow the steps mentioned in the problem to construct D, the unit digit of D always ends with the same number.

We can take advantage of this fact and construct a simple number that is 3628800(=10!) followed by 1993 0's.Then simply do what is mentioned in the problem.

P.S:- It would be an interesting exercise to prove the fact that any number with 2001 digits which is a multiple of 10!, the number D as mentioned in the problem always ends with 9.

Basically we should use the fact that the remainder of the number when divided by 9 is equal to the remainder of the sum of its digits when divided by 9 too. So as A is a multiple of 10! or 299!(most importantly a multiple of 9), it is divisible by 9, so will B hence C hence D. So D=9n for some integer n>0. Suppose we do not know is A a multiple of 10! or 299!. Maximum of B if it has 2050 digits is 2050 9=18450. So the maximum of C is 9 4=36 and hence maximum of D is 2+9=11. So D is 9 and hence its unit digit is 9. This argument should work for all other number of digit of A.

10! ends with two 0's. So A must end with a minimum of two 0's. So we actually have to find the digit sum of 2010 digits. Also note that 9 divides 10! implying that A is divisible by 9. In the worst case(might not be possible), A can have all 2010 digits from the left as 9 and remaining two as 0. In that case B=2010 9=18090. So(if possible) B can have a maximum of 5 digits and hence C can have a maximum value of 9 5= 45. We know that an integer is divisible by 9 iff the digit sum of the integer is divisible by 9. Since 9 divides A, 9 must also divide its digit sum i.e. B. Similarly 9 must also divide C. So C can have the values 09,18,27,36,45. Thus the only possible value of D is 9. Hence the unit's digit of D is 9

Take any multiple of 10! and keeping adding the digits till you get a single digit which is 9.

The value of 10!=3628800. A is a multiple of 10! which consists of 2001 digits. The smallest value of A possible is 10!×10^1994. Which is equal to 3628800.....0 in which the last digit is the 2001st. The digit sum of A is 3+6+2+8x2+0x1996 = 3+6+2+16 = 27, which is equal to B. C=2+7 = 9. D=9

The next possible value of A is 3628800...3628800 in which the last digit is the 2001st. The digit sum of A is 3x2+6x2+2x2+8x4+0x1991 = 6+12+4+32 = 54, which is equal to B. C=5+4 = 9. D=9

The next possible value of A is 3628800...7257600 in which the last digit is the 2001st. The digit sum of A is 3+6x2+2x2+8x2+7x2+5+0x1991 = 3+12+4+16+14+5 = 54 which is equal to B. C=5+4=9. D=9. Finding a pattern, the answer is always equal to 9.