Find sum of digits of D.

A= 19991999
B = sum of digits of A
C = sum of digits of B
D = sum of digits of C

This is very interesting yet I think is very easy. You just to be careful in adding the digits.

So the problem is:
A = 19991999
B = 9 * 6 = 54 + 2 = 56
C = 5 + 6 = 11
D = 1 + 1 = 2

Instead of manually adding the digits is there any way of solving this problem. Suppose we have more digits in the number manually adding would be tiring. So if some other way for doing this problem is presented it would make this discussion even more interesting.

The number 9 doens't place a value, any number added with 9 ll remains the same value in these type of cases.. so the value of D = 2.
(eg): the sum of 9197 is 26 and the sum of 2 and 6 is 8. Now cancel the 9's and add the remaining 1 7 is also 8. In these type of expressions it is better to cancel the 9's

I have a doubt...Anybody have a formula for multiplying the 9's.

My coleague give answer within a second, if i give number 9's multiplication...

Example..if i ask what is the value of 9x9x9x9x9x9 + 9x9x9x9x9x9x9 , then he told the answer within a second...

If any one have an idea about this?

sum of digits of D = E

Hi suresh,
Ans is 2