Hello,
What is the remainder when the following sum is divided by 4? $1^5 + 2^5 + 3^5 +...+ 99^5 + 100^5$
I feel like it has to do with modular arithmetic... I am trying to decompose every number but it seems to long and unnecessary. Any ideas?
P.S. thank you for your ideas. I got it. Please don't post solutions
On
hint :take {1,2,3,4},{5,6,7,8} .....{97,98,99,100} as a set now each corresponding term in each set has same remainder when divided by 4 , so you effectively need to calculate only the remainder of $1^5+2^5+3^5+4^5$ w.r.t 4 and then multiply it by 25 and again find that numbers remainder w.r.t 4
On
$$1^5 \equiv 1 \pmod{4}$$ $$4\mid 2^5 \Rightarrow 2^5 \equiv 0 \pmod{4}$$ $$3^5 \equiv (-1)^5 \equiv -1 \pmod{4}$$ $$4^5 \equiv 0^5 \equiv 0 \pmod{4}$$
This means that every even number, when raised to the fifth power, is $0 \pmod{4}$. Thus the sum you asked about is equal to the number of $1 \pmod{4}$ numbers in $\{1,2,\dots 100\}$ minus the number of $3 \pmod{4}$ numbers in $\{1,2,\dots 100\}$.
Every fourth number from $1$ to $97$, inclusive, is $1 \pmod{4}$. There are $25$ of these.
Every fourth number from $3$ to $99$, inclusive, is $3 \pmod{4}$. There are $25$ of these as well, so
$$\displaystyle\sum\limits_{n=1}^{100} n^5 \equiv 0 \pmod{4}$$
HINT : Note that in mod $4$, $$1^5\equiv1,\ \ 2^5\equiv 0,\ \ 3^5\equiv (-1)^5=-1\equiv 3,\ \ 4^5\equiv 0$$ and that $$1+0+3+0\equiv 0,\ \ 100=4\times 25.$$