If the set $$R = \{(a, b) : a + 5b = 42, a, b \in \mathbb{N}\}$$ has $$m$$ elements and $$\sum_{n=1}^{m}(1 - i^{n!}) = x + iy$$, where $$i = \sqrt{-1}$$, then the value of $$m + x + y$$ is

The elements of $$R=\{(a,b):a+5b=42,\ a,b\in\mathbb N\}$$

satisfy $$a=42-5b>0.$$

Hence,

$$5b<42\implies b=1,2,3,4,5,6,7,8.$$

Therefore,

$$m=8.$$

Now,

$$\sum_{n=1}^{m}(1-i^{n!})=8-\sum_{n=1}^{8}i^{n!}.$$

Observe that

$$i^{1!}=i,\qquad i^{2!}=i^2=-1,\qquad i^{3!}=i^6=i^2=-1.$$

Also, for $$n\ge4,$$

$$n!$$ is divisible by $$4,$$

so

$$i^{n!}=1.$$

Therefore,

$$\sum_{n=1}^{8}i^{n!}=i+(-1)+(-1)+1+1+1+1+1=3+i.$$

Hence,

$$x+iy=8-(3+i)=5-i.$$

Thus,

$$x=5,\qquad y=-1.$$

Finally,

$$m+x+y=8+5-1=12.$$

Therefore,

$$\boxed{12}.$$