Given a 2011 digit number, divisible by 9.
Let x be the sum of its digits and y the sum of the digits of x.
Find z, sum of the digits of y .
The largest 2011 digit number divisible by 9 is composed of all 9's. The sum of the digits of this number, x, is 27.
(2011 * 9) = 18099
1+8+0+9+9 = 27
The sum of the digits of 27, y, is 9.
2+7 = 9
The sum of the digit of y must be 9. Thus z is 9.
One should be aware that for any digit number divisible by 9, the sum of the digits are a multiple of 9.
The smallest ndigit positive integer divisible by 9 where z > 9 is where n is a 11111111111digit number. (2011 is only 4 digits).
x = s(11111111111 9's) = 99999999999
y = s(99999999999) = 99
z = s(99) = 18
Edited on February 21, 2012, 9:44 pm

Posted by Dej Mar
on 20120221 12:16:20 