If `A={x:x` is a multiple of 4 and x `inN}` and `B={x:x` is a multiple of 6 and x `inN}`, then `AnnB` consists of all multiples of

Options

A. 16
B. 12
C. 8
D. 4

Answer

To solve the problem, we need to find the intersection of two sets A and B, where:

- Set A consists of all natural numbers that are multiples of 4.
- Set B consists of all natural numbers that are multiples of 6.

### Step-by-Step Solution:

1. **Define Set A**:
Set A includes all multiples of 4 in the natural numbers. This can be expressed as:
\[ A = \{4, 8, 12, 16, 20, 24, 28, 32, 36, 40, \ldots\} \]

2. **Define Set B**:
Set B includes all multiples of 6 in the natural numbers. This can be expressed as:
\[ B = \{6, 12, 18, 24, 30, 36, 42, 48, 54, 60, \ldots\} \]

3. **Find the Intersection of Sets A and B**:
The intersection \( A \cap B \) consists of elements that are common to both sets A and B. We will list the elements of both sets and identify the common ones:
- From Set A: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, ...
- From Set B: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, ...

The common elements are:
- 12 (common in both)
- 24 (common in both)
- 36 (common in both)
- 48 (common in both)

Thus, we can express the intersection as:
\[ A \cap B = \{12, 24, 36, 48, \ldots\} \]

4. **Identify the Pattern**:
The elements in the intersection \( A \cap B \) are multiples of 12. This is because the least common multiple (LCM) of 4 and 6 is 12. Therefore, the intersection consists of all multiples of 12.

5. **Conclusion**:
Hence, we conclude that \( A \cap B \) consists of all multiples of 12.

### Final Answer:
\( A \cap B \) consists of all multiples of 12.

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