Question

There is a number N = 2000^11 - 2011 ; the sum of the digits of this number N is S. Then find the value of the sum of the digits of S ?

Answer 1

help please

Answer 2

Well, \(2^{11} = 2048\), and \(1000^{11} = (10^3)^{11} = 10^{33} \).

So 2000^11 would be 2048 with 33 0's on the end. Of those 33 0's, when we subtract the 2011, all but 4 of them will become 9's and the 8 in 2048 will become a 7.

Answer 3

Then we will have 10000 - 2011 = 7989 at the end of the 29 9's

Answer 4

So \[2+ 0 + 4 + 7 + 29*9 + 7 + 9 + 8 + 9 = S = 307\]
And the sum of the digits of S are 3 + 0 + 7 = 10.