If a number 18345mn is divisible by 9 then find the min vale of (m + n)?

Options

A. 8
B. 6
C. 7
D. 5

Answer

To solve the problem, we need to determine the values of \( m \) and \( n \) such that the number \( 18345mn \) is divisible by 9. According to the divisibility rule for 9, a number is divisible by 9 if the sum of its digits is divisible by 9.

### Step-by-Step Solution:

1. **Identify the digits of the number**:
The number is \( 18345mn \), which consists of the digits: 1, 8, 3, 4, 5, \( m \), and \( n \).

2. **Calculate the sum of the known digits**:
\[ 1 + 8 + 3 + 4 + 5 = 21 \]

3. **Include \( m \) and \( n \) in the sum**:
The total sum of the digits becomes:
\[ 21 + m + n \]

4. **Set up the divisibility condition**:
For the number to be divisible by 9, the sum \( 21 + m + n \) must be divisible by 9.

5. **Find the smallest multiple of 9 greater than or equal to 21**:
The multiples of 9 are 0, 9, 18, 27, 36, etc. The smallest multiple of 9 that is greater than 21 is 27.

6. **Set up the equation**:
We need:
\[ 21 + m + n = 27 \]

7. **Solve for \( m + n \)**:
Rearranging gives:
\[ m + n = 27 - 21 = 6 \]

8. **Conclusion**:
The minimum value of \( m + n \) is 6.

### Final Answer:
The minimum value of \( m + n \) is **6**.