The greatest no which shall leave equal reminders on dividing 60,132,180:

Options

A. 30
B. 36
C. 12
D. 24

Answer

To find the greatest number that leaves equal remainders when dividing 60, 132, and 180, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Remainder**:
Let the remainder when dividing the numbers be \( m \). Therefore, the numbers can be expressed as:
- \( 60 - m \)
- \( 132 - m \)
- \( 180 - m \)

2. **Formulate the Differences**:
To find a common divisor, we can calculate the differences between these numbers:
- Difference between 132 and 60:
\[ (132 - m) - (60 - m) = 132 - 60 = 72 \]
- Difference between 180 and 132:
\[ (180 - m) - (132 - m) = 180 - 132 = 48 \]

3. **Find the HCF of the Differences**:
Now, we need to find the highest common factor (HCF) of the differences 72 and 48.

4. **Factorization of 72**:
- The prime factorization of 72 is:
\[ 72 = 2^3 \times 3^2 \]

5. **Factorization of 48**:
- The prime factorization of 48 is:
\[ 48 = 2^4 \times 3^1 \]

6. **Calculate the HCF**:
- To find the HCF, we take the lowest power of each common prime factor:
- For \( 2 \): minimum power is \( 2^3 \)
- For \( 3 \): minimum power is \( 3^1 \)
- Therefore, the HCF is:
\[ HCF = 2^3 \times 3^1 = 8 \times 3 = 24 \]

7. **Conclusion**:
The greatest number that leaves equal remainders when dividing 60, 132, and 180 is **24**.