What is the remainder when the following sum is divided by 4?

[Math100]

Homework Statement

What is the remainder when the following sum is divided by $4$?
$1^{5}+2^{5}+3^{5}+\dotsb +99^{5}+100^{5}$

Relevant Equations

None.

Let $n$ be an integer.
Now we consider two cases.
Case #1: Suppose $n$ is even.
Then $n=2k$ for some $k\in\mathbb{N}$.
Thus $n^{5}=(2k)^{5}=32k^{5}\equiv 0 \pmod 4$.
Case #2: Suppose $n$ is odd.
Then $n=4k+1$ or $n=4k+3$ for some $k\in\mathbb{N}$.
Thus $(4k+1)^{5}+(4k+3)^{5}\equiv 1^{5}+(-1)^{5} \pmod 4\equiv 1+(-1) \pmod 4\equiv 0 \pmod 4$.
Note that $1^{5}+2^{5}+3^{5}+\dotsb +99^{5}+100^{5}=[(1^{5}+3^{5})+(5^{5}+7^{5})+\dotsb +(97^{5}+99^{5})]+(2^{5}+4^{5}+\dotsb +98^{5}+100^{5})\equiv (0+0) \pmod 4\equiv 0 \pmod 4$.
Therefore, the remainder of $1^{5}+2^{5}+3^{5}+\dotsb +99^{5}+100^{5}$ when divided by $4$ is $0$.

 

[fresh_42]

Homework Statement:: What is the remainder when the following sum is divided by $4$?
$1^{5}+2^{5}+3^{5}+\dotsb +99^{5}+100^{5}$
Relevant Equations:: None.

Let $n$ be an integer.
Now we consider two cases.
Case #1: Suppose $n$ is even.
Then $n=2k$ for some $k\in\mathbb{N}$.
Thus $n^{5}=(2k)^{5}=32k^{5}\equiv 0 \pmod 4$.
Case #2: Suppose $n$ is odd.
Then $n=4k+1$ or $n=4k+3$ for some $k\in\mathbb{N}$.
Thus $(4k+1)^{5}+(4k+3)^{5}\equiv 1^{5}+(-1)^{5} \pmod 4\equiv 1+(-1) \pmod 4\equiv 0 \pmod 4$.
Note that $1^{5}+2^{5}+3^{5}+\dotsb +99^{5}+100^{5}=[(1^{5}+3^{5})+(5^{5}+7^{5})+\dotsb +(97^{5}+99^{5})]+(2^{5}+4^{5}+\dotsb +98^{5}+100^{5})\equiv (0+0) \pmod 4\equiv 0 \pmod 4$.
Therefore, the remainder of $1^{5}+2^{5}+3^{5}+\dotsb +99^{5}+100^{5}$ when divided by $4$ is $0$.

I would replace the last $=$ sign by $\equiv$ to stay in the same domain $\mathbb{Z}_4$ but this is nitpicking. Your proof is fine as it is.

I have looked up the sum and it is divisible by 100. There could be more than one solution, but I don't see it immediately.

 

[pbuk]

I have looked up the sum and it is divisible by 100.

It's divisible by 2,500 but I'm not sure that helps.

There could be more than one solution, but I don't see it immediately.

There can only be one solution: the unique remainder. Do you mean more than one method for calculating the solution? You could derive the closed form for $\Sigma_1^n k^5$ but that would not be quite as easy.

 

[fresh_42]

There can only be one solution: the unique remainder. Do you mean more than one method for calculating the solution?

Sure.

You could derive the closed form for $\Sigma_1^n k^5$ but that would not be quite as easy.

I thought there must be a trick to see that 100 divides the sum without calculating too many steps.

 

[fresh_42]

$$
\dfrac{\sum_{k=1}^N k^5}{\sum_{k=1}^N k}=\dfrac{1}{6}N (N + 1) (2 N^2 + 2 N - 1)=\dfrac{1}{3} (-N^2 - N)^2 + \dfrac{1}{6} (-N^2 - N)
$$
is a nice formula. And we all know the result of $\sum_{k=1}^{100} k$ because of this anecdote:
https://en.wikipedia.org/wiki/Carl_Friedrich_Gauss#Anecdotes

 

[Delta2]

Very good proof using modulo arithmetic and the fact that an integer can be even or odd. And of course at the end the pairing of terms which reminds of the Gauss anecdote.

 

[Prof B]

No conclusion about n is drawn in Case 2. Since you are not doing a proof by cases, you really shouldn't write "Case 1" and "Case 2".

 

[fresh_42]

No conclusion about n is drawn in Case 2. Since you are not doing a proof by cases, you really shouldn't write "Case 1" and "Case 2".

I think it's ok. Maybe, he should have began with: "Let $n\in \{1,2,\ldots,100\}.$" instead of $n$ being an arbitrary integer.