In general, we have:
(99...........99)^3
p 9s
= (10^p-1)^3
= 10^(3p) - 3* 10^(2p) + 3 * 10^p -1
= 100.....00 - 300......00 + 300......00 - 1
3p 0s 2p 0s p 0s
= 99..............99700.......00299.......99
p-1 9s p-1 0s p 9s
Accordingly, the sum of the digits of (10^p -1)^3 is:
(p-1)*9 + 7 +2 + p*9
= 18*p -9 +9
= 18*p
Substituting p=2000, we have the required sum of the digits of (10^2000 - 1)^3 as:
18*2000 = 36,000
Puzzle Solution
