powers......2

Find the remainder when $1^3 + 2^3 + 3^3 + \dots + 100^3$ is divided by 6.

(1/4 n^2 (n+1)^2) mod 6 = 4

Find the remainder when $1^3 + 2^3 + 3^3 + \dots + 100^3$ is divided by 6.

$\text{The sum of } ~ 1^3 + 2^3 + 3^3 + \dots + n^3 = \left[ \frac{ n*(n+1) } {2} \right]^2\\ \text{The sum of } ~ 1^3 + 2^3 + 3^3 + \dots + 100^3 = \left[ \frac{ 100*(100+1) } {2} \right]^2$

$\begin{array}{|rcll|} \hline && 1^3 + 2^3 + 3^3 + \dots + 100^3 \pmod 6 \\ &\equiv & \left[ \frac{ 100*(100+1) } {2} \right]^2 \pmod 6 \\ &\equiv & \left[ \frac{ 100*101 } {2} \right]^2 \pmod 6 \\ &\equiv & \left[ 50*101 \right]^2 \pmod 6 \\ &\equiv & 5050^2 \pmod 6 \qquad &| \qquad 5050 \pmod 6 = 4 \\ &\equiv & 4^2 \pmod 6 \\ &\equiv & 16 \pmod 6 \\ &\equiv & 4 \pmod 6 \\ \hline \end{array}$