Sum of digits of 1999^1999 Recursively reduced to 1 digit
Answer: 1
1999^1999 = A pretty big number, so calculating this and doing is not the way.
1999 = 1998 + 1
Now all numbers with (9k+1)^n = 9M+1
Therefore 1999^1999 = 9M+1
Sum of all digits of 9M recursively would be 9
9+1 = 10
on further reduction Sum = 1
Courtesy: Meenakshi :)
29 Jun, 2011
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