To solve the equations where \( A, B, \) and \( C \) are distinct digits from 1 to 9, we rewrite the equations:
- \( A + B = C \) (1)
- \( 11A - B = 2C \) (2), since \( AA = 10A + A = 11A \)
- \( CB = 11A + A \) or \( CB = 12A \) (3), since \( AA + A = 11A + A = 12A \)
From Equation (1), we can express \( C \) in terms of \( A \) and \( B \): \[ C = A + B \] Substituting this value of \( C \) into Equation (2): \[ 11A - B = 2(A + B) \] Simplifying gives: \[ 11A - B = 2A + 2B \] Rearranging, we find: \[ 11A - 2A = 2B + B \implies 9A = 3B \implies B = 3A \text{ (4)} \]
Next, we substitute \( B = 3A \) into Equation (1): \[ C = A + B = A + 3A = 4A \text{ (5)} \]
Now substituting both expressions for \( B \) and \( C \) back into Equation (3): \[ C \cdot B = 12A \implies (4A)(3A) = 12A \] Simplifying gives us: \[ 12A^2 = 12A \] Assuming \( A \neq 0 \), we can divide both sides by \( 12A \): \[ A = 1 \]
Substituting \( A = 1 \) back to find \( B \) and \( C \): From Equation (4): \[ B = 3A = 3(1) = 3 \]
From Equation (5): \[ C = 4A = 4(1) = 4 \]
Thus, the values are:
- \( A = 1 \)
- \( B = 3 \)
- \( C = 4 \)
Now we verify these values in the original equations:
- \( A + B = 1 + 3 = 4 = C \) (valid)
- \( AA - B = 11A - B = 11 \times 1 - 3 = 11 - 3 = 8 \) and \( 2C = 2 \cdot 4 = 8 \) (valid)
- \( CB = 4 \cdot 3 = 12 \) and \( AA + A = 12A = 12 \times 1 = 12 \) (valid)
All equations are satisfied.
Thus, the sum \( A + B + C = 1 + 3 + 4 = 8 \).
The final answer is: \[ \boxed{8} \]
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