If `ABxxBA=BCB`, where A, B and C stand for just one digit and `AneBneC`, the the value of A+B+C is

Options

A. 9
B. 8
C. 6
D. 10

Answer

To solve the equation \( AB \times BA = BCB \) where \( A \), \( B \), and \( C \) are distinct digits, we will follow these steps:

### Step 1: Understand the equation
The equation \( AB \times BA \) can be interpreted as:
- \( AB \) represents the two-digit number \( 10A + B \)
- \( BA \) represents the two-digit number \( 10B + A \)
- \( BCB \) represents the three-digit number \( 100B + 10C + B = 101B + 10C \)

Thus, we can rewrite the equation as:
\[ (10A + B) \times (10B + A) = 101B + 10C \]

### Step 2: Expand the left-hand side
Now, we will expand the left-hand side:
\[ (10A + B)(10B + A) = 100AB + 10A^2 + 10B^2 + AB = 100AB + 10A^2 + 11AB + 10B^2 = 10A^2 + 111AB + 10B^2 \]

### Step 3: Set up the equation
Now we have:
\[ 10A^2 + 111AB + 10B^2 = 101B + 10C \]

### Step 4: Try values for A, B, and C
Since \( A \), \( B \), and \( C \) are digits (0-9) and must be distinct, we can start testing values. Let's try \( A = 1 \) and \( B = 2 \) as suggested in the video.

### Step 5: Substitute A and B
Substituting \( A = 1 \) and \( B = 2 \):
\[ AB = 12 \quad \text{and} \quad BA = 21 \]
Calculating \( 12 \times 21 \):
\[ 12 \times 21 = 252 \]

### Step 6: Set up for C
Now, we need to match this with \( BCB \):
\[ BCB = 101B + 10C = 101 \times 2 + 10C = 202 + 10C \]
Setting \( 252 = 202 + 10C \):
\[ 252 - 202 = 10C \implies 50 = 10C \implies C = 5 \]

### Step 7: Find A + B + C
Now we have \( A = 1 \), \( B = 2 \), and \( C = 5 \). Thus:
\[ A + B + C = 1 + 2 + 5 = 8 \]

### Final Answer
The value of \( A + B + C \) is \( 8 \).

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