Let N be the greatest number that will divide 1305 , 4665 and 6905 leaving the same remainder in each case . Then, sum of the digits in N is

Options

A. 4
B. 5
C. 6
D. 8

Answer

To solve the problem, we need to find the greatest number \( N \) that divides the numbers 1305, 4665, and 6905, leaving the same remainder in each case. We can use the property that states:

If we have three numbers \( A \), \( B \), and \( C \), the greatest number \( N \) that divides them leaving the same remainder can be found using the formula:

\[ N = \text{HCF}(B - A, C - B, C - A) \]

### Step-by-Step Solution:

1. **Identify the Numbers**:
We have three numbers:
- \( A = 1305 \)
- \( B = 4665 \)
- \( C = 6905 \)

2. **Calculate the Differences**:
We need to calculate the differences:
- \( B - A = 4665 - 1305 = 3360 \)
- \( C - B = 6905 - 4665 = 2240 \)
- \( C - A = 6905 - 1305 = 5600 \)

3. **Find the HCF**:
Now we need to find the HCF of the differences \( 3360 \), \( 2240 \), and \( 5600 \).

- **Finding HCF of 3360 and 2240**:
- Prime factorization of \( 3360 \):
\[ 3360 = 2^4 \times 3 \times 5 \times 7 \]
- Prime factorization of \( 2240 \):
\[ 2240 = 2^5 \times 5 \times 7 \]
- HCF of \( 3360 \) and \( 2240 \) is \( 2^4 \times 5 \times 7 = 1120 \).

- **Finding HCF of 1120 and 5600**:
- Prime factorization of \( 5600 \):
\[ 5600 = 2^4 \times 5^2 \times 7 \]
- HCF of \( 1120 \) and \( 5600 \) is \( 2^4 \times 5 \times 7 = 1120 \).

Thus, the HCF of \( 3360 \), \( 2240 \), and \( 5600 \) is \( 1120 \).

4. **Sum of the Digits in \( N \)**:
Now, we have \( N = 1120 \). We need to find the sum of its digits:
\[ 1 + 1 + 2 + 0 = 4 \]

### Final Answer:
The sum of the digits in \( N \) is \( 4 \).