Find sum of digits of:(1999)^1999.
[The final answer should be a single digit number, for example, (2)^16 = 65536 and the sum of its digits will be given by (6 + 5 + 5 + 3 + 6 = 25, which again will be reduced to 2 + 5 = 7].
(In reply to
super easy by Ady TZIDON)
You haven't really proved anything yet... kind of a leap of faith in several spots. Here's what Ady's trying to say I think.
First of all, when you add the digits of a certain number, you are converting all of them to ones. Conversion to ones is subtracting 9, 99, 999, 9999 from the place value, or some 1digit number times those numbers. All of these are divisible by 9, so when you add up the digits, you are just taking the number mod 9.
Since any number 1 mod 9 multiplied by another number 1 mod 9 is also 1 mod 9, the answer is 1. This is because if you express these two numbers (where x and y are integers) as 9x+1 and 9y+1, you get 81xy+9x+9y+1, or 9(9xy+x+y)+1, which is clearly 1 mod 9.

Posted by Gamer
on 20040127 09:06:13 