# Thread: remainder of sum, divided by 4

1. ## remainder of sum, divided by 4

Ok, so I did the problem and got a solution that another classmate didn't get. Now its got me paranoid, I don't know whose right.

"What is the remainder when the following sum is divided by 4?": $1^5 + 2^5 + 3^5 + ... + 99^5 + 100^5$

So I'll briefly explain - I did:
so that is $\equiv 25(1^5 + 2^5 + 3^5 + 0^5)$ (mod 4)
worked it all out so that it is now congruent to 1(1 + 0 + 0) (mod 4)
Thus, the remainder is 1, yes??

My classmate go that there was no remainder. Did I do something wrong in calculations??

2. Originally Posted by cassiopeia1289
Ok, so I did the problem and got a solution that another classmate didn't get. Now its got me paranoid, I don't know whose right.

"What is the remainder when the following sum is divided by 4?": $1^5 + 2^5 + 3^5 + ... + 99^5 + 100^5$

So I'll briefly explain - I did:
so that is $\equiv 25(1^5 + 2^5 + 3^5 + 0^5)$ (mod 4)
worked it all out so that it is now congruent to 1(1 + 0 + 0) (mod 4)
Thus, the remainder is 1, yes??

My classmate go that there was no remainder. Did I do something wrong in calculations??
Your idea is very good, but be careful, $3^5\equiv(-1)^5\equiv -1\equiv 3\,({\rm mod}\,4)$ hence $1^5+2^5+3^5+4^5\equiv 1+0+3+0\equiv 0\,({\rm mod}\,4)$, and your classmate is right.

By the way, $3^5\equiv 0\,({\rm mod}\, 4)$ would mean that 4 divides $3^5$. Of course, this can't be.

3. oh ok - thanks! I knew it was something stupid I did!