Number theory

What is the remainder of $N=1\times 3\times5\times7\times...\times101$  when it is divided by 8?

So
$\begin{array}{rcll} N &\equiv& x \pmod{8} \\ 1\times 3\times5\times7\times\ldots\times101 &\equiv& x \pmod{8} \\ \end{array}$

$\begin{array}{|rcl|rcl|rcl|rcl|} \hline \mathbf{1} &\equiv& 1 \pmod{8} & \mathbf{3} &\equiv& 3 \pmod{8}& \mathbf{5} &\equiv& 5 \pmod{8}& \mathbf{7} &\equiv& 7 \pmod{8} \\ \mathbf{9} &\equiv& 1 \pmod{8} & \mathbf{11} &\equiv& 3 \pmod{8}& \mathbf{13} &\equiv& 5 \pmod{8}& \mathbf{15} &\equiv& 7 \pmod{8} \\ \mathbf{17} &\equiv& 1 \pmod{8} & \mathbf{19} &\equiv& 3 \pmod{8}& \mathbf{21} &\equiv& 5 \pmod{8}& \mathbf{23} &\equiv& 7 \pmod{8} \\ \mathbf{25} &\equiv& 1 \pmod{8} & \mathbf{27} &\equiv& 3 \pmod{8}& \mathbf{29} &\equiv& 5 \pmod{8}& \mathbf{31} &\equiv& 7 \pmod{8} \\ \vdots &&&\vdots&&&\vdots&&&\vdots \\ \mathbf{89} &\equiv& 1 \pmod{8} & \mathbf{91} &\equiv& 3 \pmod{8}& \mathbf{93} &\equiv& 5 \pmod{8}& \mathbf{95} &\equiv& 7 \pmod{8} \\ \mathbf{87} &\equiv& 1 \pmod{8} & \mathbf{99} &\equiv& 3 \pmod{8}& \mathbf{101} &\equiv& 5 \pmod{8} \\ \hline \end{array}$

$\small{ \begin{array}{|rcll|} \hline (1\times 3\times5\times7)\times (9\times11\times13\times 15)\times (17\times19\times21\times 23)\times \ldots \times (89\times91\times93\times 95)\times(97 \times99 \times101) &\equiv& x \pmod{8} \\ (1\times 3\times5\times7)\times (1\times 3\times5\times7)\times (1\times 3\times5\times7)\times \ldots \times (1\times 3\times5\times7)\times(1 \times3 \times5) &\equiv& x \pmod{8} \\ \boxed{ (1\times 3\times5\times7) = 105 \equiv 1\pmod{8}\\ (1\times 3\times5) = 15 \equiv 7\pmod{8} } \\ (1)\times (1)\times (1)\times \ldots \times (1)\times(7) &\equiv& {\color{red}7} \pmod{8} \\ \hline \end{array} }$

The remainder of $N=1\times 3\times5\times7\times\cdots\times101$  when it is divided by 8 is $\mathbf{\color{red}7}$