$$R=\{(a, b): a+5 b=42\}$$

Then $$R=\{(2,8),(7,7),(12,6),(17,5),(22,4),(27, 3),(32,2),(37,1)\}$$

$$\begin{aligned} & \text { and } \sum_{n=1}^{\substack{m=8}}\left(1-i^{n!}\right)=x+i y \\ & \therefore \sum_{n=1}^8\left(1-i^{n!}\right)=8-\left(i+i^2+i^6+1+1+1+1+1\right) \\ & =5-i \\ & \therefore x=5, y=-1 \\ & x+y+m=5-1+8=12 \end{aligned}$$