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

Options

A. N
B. `N_(5)`
C. `N_(7)`
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-by-Step Solution:

1. **Define the Sets**:
- \( N_5 = \{ 5n : n \in \mathbb{N} \} = \{ 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, \ldots \} \)
- \( N_7 = \{ 7n : n \in \mathbb{N} \} = \{ 7, 14, 21, 28, 35, 42, 49, 56, \ldots \} \)

2. **Identify the Elements of Each Set**:
- The elements of \( N_5 \) are all multiples of 5.
- The elements of \( N_7 \) are all multiples of 7.

3. **Find the Intersection \( N_5 \cap N_7 \)**:
- The intersection of two sets consists of elements that are common to both sets.
- We need to find common multiples of 5 and 7. The least common multiple (LCM) of 5 and 7 is 35.

4. **List the Common Elements**:
- The common elements will be multiples of 35, since 35 is the smallest number that is a multiple of both 5 and 7.
- Thus, the intersection \( N_5 \cap N_7 = \{ 35n : n \in \mathbb{N} \} = \{ 35, 70, 105, 140, \ldots \} \)

5. **Conclusion**:
- Therefore, \( N_5 \cap N_7 \) is the set of all multiples of 35.

### Final Answer:
\( N_5 \cap N_7 = \{ 35n : n \in \mathbb{N} \} \)

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