If 1A2B5 is exactly divisible by 9, then the least value of (A+B) is

Options

A. 0
B. 1
C. 2
D. 10

Answer

To determine the least value of \( A + B \) such that the number \( 1A2B5 \) is exactly divisible by 9, we can follow these steps:

### Step 1: Understand the divisibility rule for 9
A number is divisible by 9 if the sum of its digits is divisible by 9.

### Step 2: Calculate the sum of the known digits
The digits of the number \( 1A2B5 \) are \( 1, A, 2, B, 5 \). We can calculate the sum of the known digits:
\[ 1 + 2 + 5 = 8 \]

### Step 3: Include \( A \) and \( B \) in the sum
Now, we can express the total sum of the digits as:
\[ 8 + A + B \]

### Step 4: Set up the condition for divisibility by 9
For \( 1A2B5 \) to be divisible by 9, the total sum \( 8 + A + B \) must be divisible by 9. We can express this as:
\[ 8 + A + B \equiv 0 \mod 9 \]

### Step 5: Find the smallest value of \( A + B \)
To find the smallest value of \( A + B \), we can calculate:
\[ A + B \equiv 1 \mod 9 \]
This means that \( A + B \) can take values such as \( 1, 10, 19, \ldots \). However, since \( A \) and \( B \) are single-digit numbers (0 to 9), the only feasible solution is:
\[ A + B = 1 \]

### Step 6: Determine possible values for \( A \) and \( B \)
The pairs of \( (A, B) \) that satisfy \( A + B = 1 \) are:
- \( A = 0, B = 1 \)
- \( A = 1, B = 0 \)

### Conclusion
The least value of \( A + B \) is \( 1 \).