If the HCF of three numbers 144, x and 192 is 12, then the number x cannot be ?

Options

A. 180
B. 84
C. 60
D. 48

Answer

To solve the problem, we need to determine which number \( x \) cannot be, given that the HCF (Highest Common Factor) of the three numbers 144, \( x \), and 192 is 12.

### Step 1: Find the prime factorization of the numbers

1. **Prime factorization of 144:**
\[ 144 = 12 \times 12 = (2^2 \times 3) \times (2^2 \times 3) = 2^4 \times 3^2 \]

2. **Prime factorization of 192:**
\[ 192 = 96 \times 2 = (2^5 \times 3) \times 2 = 2^6 \times 3^1 \]

### Step 2: Determine the HCF condition

Given that the HCF of 144, \( x \), and 192 is 12, we can express 12 in terms of its prime factors:
\[ 12 = 2^2 \times 3^1 \]

### Step 3: Analyze the HCF with respect to \( x \)

For the HCF of the three numbers to be 12, \( x \) must have at least the prime factors \( 2^2 \) and \( 3^1 \). Therefore, \( x \) can be expressed as:
\[ x = 2^a \times 3^b \times k \]
where \( a \geq 2 \), \( b \geq 1 \), and \( k \) is any integer that does not introduce any additional common factors with 144 or 192.

### Step 4: Identify the constraints on \( x \)

Since we want to find a value of \( x \) that cannot satisfy the HCF condition, we need to ensure that \( x \) does not meet the required prime factorization conditions.

### Step 5: Test possible values for \( x \)

Let’s consider some values for \( x \):

1. **If \( x = 48 \):**
\[ 48 = 2^4 \times 3^1 \]
The HCF of \( 144, 48, 192 \) would be:
\[ HCF(2^4 \times 3^2, 2^4 \times 3^1, 2^6 \times 3^1) = 2^4 \times 3^1 = 48 \quad (\text{not } 12) \]

2. **If \( x = 60 \):**
\[ 60 = 2^2 \times 3^1 \times 5 \]
The HCF would be:
\[ HCF(2^4 \times 3^2, 2^2 \times 3^1 \times 5, 2^6 \times 3^1) = 2^2 \times 3^1 = 12 \]

3. **If \( x = 84 \):**
\[ 84 = 2^2 \times 3^1 \times 7 \]
The HCF would be:
\[ HCF(2^4 \times 3^2, 2^2 \times 3^1 \times 7, 2^6 \times 3^1) = 2^2 \times 3^1 = 12 \]

### Conclusion

From the analysis, we find that \( x = 48 \) does not satisfy the condition that the HCF of 144, \( x \), and 192 equals 12. Therefore, the number \( x \) cannot be 48.