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 1-digit 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 2004-01-27 09:06:13