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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