Benedict Irwin edited Rules.tex  almost 9 years ago

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\end{array} \\  \frac{Z}{11} \in \mathbb{Z}  \end{equation}  Or rather more succinctly as: If \begin{equation}  (a+b-3)\; \mathrm{mod} \; 11 =0 \wedge n-1 \; \mathrm{mod} \; 2 = 0  \end{equation}  \section{13}  For divisibility by $13$, the general $\Gamma$ matrix can be constructed below 

\end{equation}  Here we can see that if a number occurs, then that number plus $6$ also occurs if it is still in the range modulo $[1,12]$.  \begin{equation}  \Theta_{1}= \Theta_{1}\to\Theta_{7}=  \begin{bmatrix}  1 & & & & & & & & & \\  & & & &1 & & & & & \\ 

\end{bmatrix}  \end{equation}  \begin{equation}  \Theta_{2}= \Theta_{2}\to\Theta{8}=  \begin{bmatrix}  & &2 & & & & & & & \\  & & &2 & & & & & & \\ 

\end{bmatrix}  \end{equation}  \begin{equation}  \Theta_{4}= \Theta_{4}\to\Theta{10}=  \begin{bmatrix}  & & & & & & & & & \\  & & & & & & & &4 & \\ 

\end{bmatrix}  \end{equation}  Leaving Rules  \begin{equation}  \begin{array}  \; & [(a+3b) & mod \; 13 =0 & \wedge & n - 1 \; mod \; 6 = 0] \\ % 1 & 7  \vee & [(b-a-1) & mod \; 13 =0 & \wedge & n - 2 \; mod \; 6 = 0] \\ % 2 & 8  \vee & [(a+b-3) & mod \; 13 =0 & \wedge & n - 5 \; mod \; 6 = 0] \\ % 5 & 11  \end{array} \\  \frac{Z}{13} \in \mathbb{Z}  \end{equation}  \section{17}  For divisibility by $17$, the general $\Gamma$ matrix can be constructed below  \begin{equation}