If `N_(a)={an:ninN}`, then `N_(5) nn N_(7)` equals

Options

A. A. N
B. B. `N_(5)`
C. C. `N_(7)`
D. D. `N_(35)`

Answer

To solve the problem, we need to find the intersection of the sets \( N_5 \) and \( N_7 \), where:

- \( N_a = \{ an : n \in \mathbb{N} \} \)

This means that \( N_5 \) consists of all multiples of 5, and \( N_7 \) consists of all multiples of 7.

### Step 1: Define the sets \( N_5 \) and \( N_7 \)

1. **Finding \( N_5 \)**:
\[ N_5 = \{ 5n : n \in \mathbb{N} \} = \{ 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, \ldots \} \]

2. **Finding \( N_7 \)**:
\[ N_7 = \{ 7n : n \in \mathbb{N} \} = \{ 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, \ldots \} \]

### Step 2: Find the intersection \( N_5 \cap N_7 \)

To find the intersection \( N_5 \cap N_7 \), we need to identify the common elements in both sets. The common elements will be the multiples of both 5 and 7.

### Step 3: Determine the least common multiple (LCM)

The least common multiple of 5 and 7 is:
\[ \text{lcm}(5, 7) = 35 \]

### Step 4: Express the intersection in terms of \( N \)

The intersection can be expressed as:
\[ N_5 \cap N_7 = \{ 35n : n \in \mathbb{N} \} = N_{35} \]

### Conclusion

Thus, the intersection \( N_5 \cap N_7 \) is equal to \( N_{35} \).

### Final Answer
The answer is \( N_{35} \).

---