Question

If A, B, and C represent three distinct digits from 1 to 9 and they satisfy the following equations, what is the value of the sum A + B + C? (In the equation below, AA represents a two-digit number both of whose digits are A)

A + B = C
AA - B = 2C
C*B = AA + A

Answer 1

To solve the equations where \( A, B, \) and \( C \) are distinct digits from 1 to 9, we rewrite the equations:

1. \( A + B = C \) (1)
2. \( 11A - B = 2C \) (2), since \( AA = 10A + A = 11A \)
3. \( CB = 11A + A \) 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:
1. \( A + B = 1 + 3 = 4 = C \) (valid)
2. \( AA - B = 11A - B = 11 \times 1 - 3 = 11 - 3 = 8 \) and \( 2C = 2 \cdot 4 = 8 \) (valid)
3. \( CB = 4 \cdot 3 = 12 \) 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:
\[
\boxed{8}
\]