Assuming A ,B and C are different single digit numerical values other than what is already used in following equation. What number C definitely cannot be ?
8A2 + 3B5 + C4 = 1271

Options

A. 7
B. 9
C. Either 7 or 9
D. 6

Answer

To solve the equation \( 8A2 + 3B5 + C4 = 1271 \) and determine which digit \( C \) definitely cannot be, we will follow these steps:

### Step 1: Understand the equation
The equation consists of three terms: \( 8A2 \), \( 3B5 \), and \( C4 \). Here, \( A \), \( B \), and \( C \) are single-digit numbers (0-9) that must be different from each other and cannot be any digit already used in the equation.

### Step 2: Identify used digits
The digits in the equation are:
- From \( 8A2 \): 8, 2
- From \( 3B5 \): 3, 5
- From \( C4 \): 4

Thus, the digits already used are: 2, 3, 4, 5, and 8.

### Step 3: List possible digits for A, B, and C
The digits that can be used for \( A \), \( B \), and \( C \) are those not already used. The digits available are:
- 0, 1, 6, 7, 9

### Step 4: Check the constraints
Since \( A \), \( B \), and \( C \) must be different, we need to check which of the available digits can be assigned to \( C \) without conflicting with \( A \) and \( B \).

### Step 5: Calculate the total contribution of known digits
Let's calculate the contribution of the known digits:
- \( 8A2 = 800 + 10A + 2 = 802 + 10A \)
- \( 3B5 = 300 + 10B + 5 = 305 + 10B \)
- \( C4 = 10C + 4 \)

Combining these, we have:
\[ 802 + 10A + 305 + 10B + 10C + 4 = 1271 \]
This simplifies to:
\[ 10A + 10B + 10C + 1111 = 1271 \]
\[ 10A + 10B + 10C = 160 \]
\[ A + B + C = 16 \]

### Step 6: Analyze possible values for C
Now we need to find combinations of \( A \), \( B \), and \( C \) that add up to 16, using the available digits (0, 1, 6, 7, 9).

- If \( C = 0 \): \( A + B = 16 \) (not possible with single digits)
- If \( C = 1 \): \( A + B = 15 \) (not possible with single digits)
- If \( C = 6 \): \( A + B = 10 \) (possible combinations: 1+9, 2+8, 3+7, 4+6)
- If \( C = 7 \): \( A + B = 9 \) (possible combinations: 0+9, 1+8, 2+7, 3+6, 4+5)
- If \( C = 9 \): \( A + B = 7 \) (possible combinations: 0+7, 1+6, 2+5, 3+4)

### Step 7: Determine which digit C cannot be
From the analysis, we see that:
- \( C \) cannot be 0 or 1 as they do not allow valid combinations for \( A \) and \( B \).
- \( C \) can be 6, 7, or 9.

However, since we are looking for a digit that \( C \) definitely cannot be, we see that:
- \( C \) cannot be 7 because if \( C = 7 \), then \( A + B = 9 \) could lead to a conflict with the digits already used (3, 5, 8).

### Conclusion
Thus, the digit that \( C \) definitely cannot be is **7**.