1. Find the number of elements in $R$

We need the number of natural number solutions of $a+5b=42,$ with $a,b\in\mathbb N$.

Assuming $\mathbb N=\{1,2,3,\dots\}$, we write $a=42-5b.$ Since $a\in\mathbb N$, we need $42-5b\ge 1 \implies 5b\le 41 \implies b\le 8.2.$ So $b=1,2,3,4,5,6,7,8.$ That gives $m=8.$

2. Evaluate $i^{n!}$

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

Recall powers of $i$ repeat modulo $4$: $i^0=1,\quad i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1.$

Now compute term by term:

• For $n=1$: $1!=1$, so $i^{1!}=i.$ Term is $1-i.$

• For $n\ge 2$: $n!$ is divisible by $2$, and for $n\ge 4$ it is certainly divisible by $4$. In fact: $2!=2 \Rightarrow i^{2!}=i^2=-1,$ $3!=6 \Rightarrow i^{3!}=i^6=i^2=-1,$ and for $n=4,5,6,7,8$, since $n!\equiv 0\pmod 4$, $i^{n!}=1.$

So the sum becomes $S=(1-i)+(1-(-1))+(1-(-1))+(1-1)+(1-1)+(1-1)+(1-1)+(1-1).$

That is $S=(1-i)+2+2=5-i.$ Hence $x=5,\quad y=-1.$

3. Compute $m+x+y$

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

4. Compare with options

The value is $\boxed{12}.$ So the correct option is A.

5. Check with stored answer

Stored correct answer: A

Our derived answer also gives A. Hence they agree.