<s>[INST] Find remainder when $1^{5} + 2^{5} \cdots +100^{5}$ divided by 4 I'm studding D.M Burton & want to solve:  Find remainder when $1^{5} + 2^{5} \cdots +100^{5}$ divided by $4$. . Please help me by giving your solution to it. I'm new comer to number theory so please don't use theorems above Theory of Congruence.
 [/INST] You can do it even without knowledge of congruences, we have that $1^5 + 2^5 + ... + n^5= \dfrac {(n(n+1))^2(2n^2+2n-1)}{12}$, now set $n=100$ to get $1^5 + 2^5 + ... + 100^5 = \dfrac {(100 \cdot 101)^2 (2 \cdot 100^2 + 2 \cdot 100 - 1)}{12}=4 \cdot \dfrac {4 \cdot 25^2 \cdot 101^2 \cdot 3 \cdot 6733}{12}$ so your number is divisible by $4$.
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