+0

# A congruence

0
203
4

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.

Nov 27, 2018

#1
+770
0

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.

Nov 28, 2018
#2
+770
0

I'll probably feed the data into a computer program to determine the answer.

- PM

PartialMathematician  Nov 28, 2018
#3
+6187
0

It's a bit more clever than that.

If you look at k^1000 % 1000 you'll see it's a periodic sequence with period 10

I'm no number theory guru but I think the secret lies in the common factors of k and 1000.

I'll post if I figure it out.

Rom  Nov 28, 2018
#4
0

I don't know if there is a shortcut to it, but computers can easily handle it. It is 2001-digit long.
It begins with......10000431729.......and ends in....  5137781330 mod 1000 =330

Nov 28, 2018