Puzzle 2: Find sum of digits of D.

Find sum of digits of D.
Let
A= 19991999
B = sum of digits of A
C = sum of digits of B
D = sum of digits of C

(HINT : A = B = C = D (mod 9))

Answer:

The sum of the digits od D is 1.

Let E = sum of digits of D.
It follows from the hint that A = E (mod 9)
Consider,
A = 19991999
< 20002000 =" 22000" 10002000 =" 1024200" 106000 =" 106800" 9 =" 61200" 9 =" 45" 9 =" 18" 1999 =" 1" 19991999 =" 1" e="1.