The greatest number which divides 261, 933 and 1381 leaving remainders 5 in each case is

Options

A. 128
B. 64
C. 32
D. 16

Answer

To find the greatest number which divides 261, 933, and 1381 leaving a remainder of 5 in each case, we can follow these steps:

### Step 1: Subtract the remainder from each number
Since we want to find a number that leaves a remainder of 5, we first subtract 5 from each of the given numbers.

- \( 261 - 5 = 256 \)
- \( 933 - 5 = 928 \)
- \( 1381 - 5 = 1376 \)

### Step 2: Find the HCF of the new numbers
Now, we need to find the highest common factor (HCF) of the three new numbers: 256, 928, and 1376.

### Step 3: Prime factorization of each number
1. **Prime factorization of 256**:
- \( 256 = 2^8 \)

2. **Prime factorization of 928**:
- \( 928 = 2^4 \times 58 \)
- Further factorizing 58 gives \( 58 = 2 \times 29 \)
- Thus, \( 928 = 2^5 \times 29 \)

3. **Prime factorization of 1376**:
- \( 1376 = 2^4 \times 86 \)
- Further factorizing 86 gives \( 86 = 2 \times 43 \)
- Thus, \( 1376 = 2^5 \times 43 \)

### Step 4: Identify the common factors
Now, we look for the common factors in the prime factorizations:
- \( 256 = 2^8 \)
- \( 928 = 2^5 \times 29 \)
- \( 1376 = 2^5 \times 43 \)

The common factor among these numbers is \( 2^5 \).

### Step 5: Calculate the HCF
The HCF is \( 2^5 = 32 \).

### Step 6: Conclusion
The greatest number which divides 261, 933, and 1381 leaving a remainder of 5 in each case is **32**.

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