Set `A = {(a, b) : a, b in N`, `a + 5b = 42}` has m elements also `sum_(n=1)^m(1 – i^(n!)) = x + iy`, `i = sqrt -1`, then `m + x + y`

Answer

To solve the problem step by step, we will break it down into manageable parts.

### Step 1: Determine the set A
We start with the equation given for the set A:
\[ A = \{(a, b) : a, b \in \mathbb{N}, a + 5b = 42\} \]

From this equation, we can express \( a \) in terms of \( b \):
\[ a = 42 - 5b \]

Since \( a \) and \( b \) must be natural numbers (positive integers), we need to find the possible values of \( b \) such that \( a \) remains a natural number.

### Step 2: Find the values of b
To ensure \( a \) is a natural number, \( 42 - 5b > 0 \):
\[ 42 > 5b \]
\[ b < \frac{42}{5} = 8.4 \]

Since \( b \) must be a natural number, the possible values for \( b \) are \( 1, 2, 3, 4, 5, 6, 7, 8 \).

### Step 3: Calculate corresponding values of a
Now, we can calculate the corresponding values of \( a \) for each \( b \):
- For \( b = 1 \): \( a = 42 - 5(1) = 37 \) → (37, 1)
- For \( b = 2 \): \( a = 42 - 5(2) = 32 \) → (32, 2)
- For \( b = 3 \): \( a = 42 - 5(3) = 27 \) → (27, 3)
- For \( b = 4 \): \( a = 42 - 5(4) = 22 \) → (22, 4)
- For \( b = 5 \): \( a = 42 - 5(5) = 17 \) → (17, 5)
- For \( b = 6 \): \( a = 42 - 5(6) = 12 \) → (12, 6)
- For \( b = 7 \): \( a = 42 - 5(7) = 7 \) → (7, 7)
- For \( b = 8 \): \( a = 42 - 5(8) = 2 \) → (2, 8)

Thus, the set \( A \) has 8 elements:
\[ A = \{(37, 1), (32, 2), (27, 3), (22, 4), (17, 5), (12, 6), (7, 7), (2, 8)\} \]

So, \( m = 8 \).

### Step 4: Calculate the sum
Next, we need to evaluate the sum:
\[ \sum_{n=1}^{m} (1 - i^{n!}) \]

Substituting \( m = 8 \):
\[ \sum_{n=1}^{8} (1 - i^{n!}) \]

Calculating \( i^{n!} \) for \( n = 1, 2, \ldots, 8 \):
- \( n = 1 \): \( i^{1!} = i^1 = i \)
- \( n = 2 \): \( i^{2!} = i^2 = -1 \)
- \( n = 3 \): \( i^{3!} = i^6 = (i^4)(i^2) = 1 \cdot (-1) = -1 \)
- \( n = 4 \): \( i^{4!} = i^{24} = (i^4)^6 = 1^6 = 1 \)
- \( n = 5 \): \( i^{5!} = i^{120} = (i^4)^{30} = 1^{30} = 1 \)
- \( n = 6 \): \( i^{6!} = i^{720} = (i^4)^{180} = 1^{180} = 1 \)
- \( n = 7 \): \( i^{7!} = i^{5040} = (i^4)^{1260} = 1^{1260} = 1 \)
- \( n = 8 \): \( i^{8!} = i^{40320} = (i^4)^{10080} = 1^{10080} = 1 \)

Now substituting these values back into the sum:
\[ \sum_{n=1}^{8} (1 - i^{n!}) = (1 - i) + (1 - (-1)) + (1 - (-1)) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) \]
This simplifies to:
\[ (1 - i) + 2 + 0 + 0 + 0 + 0 + 0 + 0 = 3 - i \]

### Step 5: Identify x and y
From the sum \( 3 - i \), we identify:
- \( x = 3 \)
- \( y = -1 \)

### Step 6: Calculate \( m + x + y \)
Now, we can calculate:
\[ m + x + y = 8 + 3 - 1 = 10 \]

### Final Answer
The final answer is:
\[ \boxed{10} \]