What is the remainder of the following congruence:
[1^1000 + 2^1000 + 3^1000 +..................+100^1000] mod 1000 =? Any help would be appreciated. Thank you.
+773 This is a really tedious number theory problem. I would probably try to find a pattern in the mod 1000 of the powers of 2, 3, 4, ect. Remember that any multiple of 10 to the power of 1000 (\(10^{1000}\), \(20^{1000}\), ect.) mod 1000 is 0, which means it can be divided by 1000. Other than that, I think there is just pure computations for this problem.
+773 I'll probably feed the data into a computer program to determine the answer.
- PM
