Question: 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
< 10800 * 106000
= 106800
i.e. A < 106800
i.e. B < 6800 * 9 = 61200
i.e. C < 5 * 9 = 45
i.e. D < 2 * 9 = 18
i.e. E <= 9
i.e. E is a single digit number.
Also,
1999 = 1 (mod 9)
so 19991999 = 1 (mod 9)
Therefore we conclude that E=1.