a, b, c are digits of a 3-digit number such that 64a + 8b + c = 403, then the value of a + b + c + 2013 is

Answer

To solve the problem, we need to find the values of the digits \( a \), \( b \), and \( c \) from the equation:

\[ 64a + 8b + c = 403 \]

### Step 1: Simplify the Equation
First, we can observe that the entire equation is a multiple of 8. Therefore, we can divide the entire equation by 8:

\[ \frac{64a}{8} + \frac{8b}{8} + \frac{c}{8} = \frac{403}{8} \]

This simplifies to:

\[ 8a + b + \frac{c}{8} = 50 + \frac{3}{8} \]

### Step 2: Separate the Integer and Fractional Parts
From the equation above, we can separate the integer and fractional parts:

\[ 8a + b = 50 \quad \text{and} \quad c = 3 \]

### Step 3: Solve for \( a \) and \( b \)
Now we know \( c = 3 \). We need to find suitable values for \( a \) and \( b \) such that \( 8a + b = 50 \). Since \( a \) and \( b \) are digits (0-9), we can try different values for \( a \):

- If \( a = 6 \):
\[ 8 \times 6 + b = 50 \implies 48 + b = 50 \implies b = 2 \]

- If \( a = 7 \):
\[ 8 \times 7 + b = 50 \implies 56 + b = 50 \implies b < 0 \quad (\text{not valid}) \]

Thus, the only valid solution is \( a = 6 \), \( b = 2 \), and \( c = 3 \).

### Step 4: Calculate \( a + b + c + 2013 \)
Now we can find the required sum:

\[ a + b + c + 2013 = 6 + 2 + 3 + 2013 = 11 + 2013 = 2024 \]

### Final Answer
The final answer is:

\[ \boxed{2024} \]

---