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)

Question

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: $$11A-2A=2B+B⟹9A=3B⟹B=3A\text{ (4)}$$

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): $$C\cdot B=12A⟹(4A)(3A)=12A$$ Simplifying gives us: $$12A^2=12A$$ Assuming $A\ne 0$, 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: $$\boxed{{\displaystyle 8}}$$