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.

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