The numbers 2272 and 875 are divided by a 3-digit number N, giving the same remaindes. The sum of the digits of N is

Options

A. 10
B. 11
C. 12
D. 13

Answer

To solve the problem, we need to find a 3-digit number \( N \) such that when both 2272 and 875 are divided by \( N \), they yield the same remainder.

### Step-by-step Solution:

1. **Understanding the Problem**:
We are looking for a 3-digit number \( N \) such that:
\[ 2272 \mod N = 875 \mod N \]
This implies that the difference between the two numbers, \( 2272 - 875 \), must be divisible by \( N \).

2. **Calculate the Difference**:
First, we calculate the difference:
\[ 2272 - 875 = 1397 \]

3. **Finding Divisors of the Difference**:
Now, we need to find the divisors of 1397. We will check if 1397 is divisible by any 3-digit number.

4. **Factorizing 1397**:
We can start by checking if 1397 is divisible by small prime numbers:
- 1397 is not even, so it is not divisible by 2.
- The sum of the digits \( 1 + 3 + 9 + 7 = 20 \) is not divisible by 3, so it is not divisible by 3.
- It does not end in 0 or 5, so it is not divisible by 5.
- Checking for divisibility by 7, 11, 13, etc., we find that:
\[ 1397 = 127 \times 11 \]

5. **Identifying the 3-Digit Number**:
From the factorization, we see that 1397 can be expressed as \( 127 \times 11 \). Since we are looking for a 3-digit number, we can take \( N = 127 \) (which is a 3-digit number).

6. **Sum of the Digits of N**:
Now we calculate the sum of the digits of \( N \):
\[ 1 + 2 + 7 = 10 \]

### Final Answer:
Thus, the sum of the digits of \( N \) is \( 10 \).