If the least common multiple of two numbers, 1728 and K is 5184, then how many values of K are possible?

Options

A. 11
B. 8
C. 6
D. 7

Answer

To solve the problem, we need to find how many values of \( K \) are possible such that the least common multiple (LCM) of \( 1728 \) and \( K \) is \( 5184 \).

### Step-by-Step Solution:

1. **Prime Factorization of 1728**:
- First, we perform the prime factorization of \( 1728 \).
- \( 1728 = 12^3 = (2^2 \cdot 3)^3 = 2^6 \cdot 3^3 \).
- Therefore, the prime factorization of \( 1728 \) is:
\[ 1728 = 2^6 \cdot 3^3 \]

2. **Prime Factorization of 5184**:
- Next, we perform the prime factorization of \( 5184 \).
- \( 5184 = 2^6 \cdot 3^4 \).
- Therefore, the prime factorization of \( 5184 \) is:
\[ 5184 = 2^6 \cdot 3^4 \]

3. **Understanding LCM**:
- The LCM of two numbers is found by taking the highest power of each prime factor present in either number.
- From the prime factorizations:
- For \( 2 \): The maximum power is \( 2^6 \) (from both \( 1728 \) and \( 5184 \)).
- For \( 3 \): The maximum power is \( 3^4 \) (from \( 5184 \)).

4. **Finding the Required Form of K**:
- Since \( \text{LCM}(1728, K) = 5184 \), we can express \( K \) in terms of its prime factors:
\[ K = 2^a \cdot 3^b \]
- Where \( a \) and \( b \) must satisfy:
- \( \max(6, a) = 6 \) (which means \( a \) can be \( 0, 1, 2, 3, 4, 5, \) or \( 6 \)).
- \( \max(3, b) = 4 \) (which means \( b \) must be \( 4 \) since \( b \) must be greater than \( 3 \)).

5. **Determining Values of K**:
- From the above conditions:
- \( a \) can take any value from \( 0 \) to \( 6 \) (7 possible values).
- \( b \) must be \( 4 \) (1 possible value).
- Therefore, the total number of possible values of \( K \) is:
\[ 7 \text{ (values for } a) \times 1 \text{ (value for } b) = 7 \]

### Conclusion:
Thus, the number of possible values of \( K \) is \( 7 \).