**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.