The L.C.M. of three different numbers is 120 . Which of the follosing cannot be their H.C.F. ?

Options

A. 8
B. 12
C. 24
D. 35

Answer

To solve the problem, we need to determine which of the given options cannot be the H.C.F. (Highest Common Factor) of three different numbers whose L.C.M. (Lowest Common Multiple) is 120.

### Step-by-Step Solution:

1. **Understand the relationship between LCM and HCF**:
The relationship between the LCM and HCF of any two or more numbers is given by the formula:
\[ \text{LCM} \times \text{HCF} = \text{Product of the numbers} \]
Since we have three numbers, the same principle applies.

2. **Factorize the LCM**:
We start by factorizing the LCM, which is 120.
\[ 120 = 2^3 \times 3^1 \times 5^1 \]

3. **Identify the possible HCF**:
The HCF must be a divisor of the LCM. Therefore, we need to find the divisors of 120. The divisors of 120 can be calculated from its prime factorization:
- The divisors of 120 are: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120.

4. **Check the given options**:
The options provided are:
- 8
- 12
- 24
- 35

5. **Determine which options are divisors of 120**:
- **8**: Yes, 8 is a divisor of 120.
- **12**: Yes, 12 is a divisor of 120.
- **24**: Yes, 24 is a divisor of 120.
- **35**: No, 35 is not a divisor of 120.

6. **Conclusion**:
Since 35 is not a divisor of 120, it cannot be the HCF of three numbers whose LCM is 120. Therefore, the answer is:
\[ \text{Option 4: 35 cannot be their HCF.} \]