Let $ABC$ represent a three-digit number where $A$, $B$, and $C$ are its digits.

In terms of place values, the number $ABC$ can be written as: $100A + 10B + C$ The reversed number $CBA$ is: $100C + 10B + A$

Given Conditions

1. $A + B + C = 12$
2. $CBA = ABC - 594$

Step 1: Set Up an Equation for Condition 2

According to condition 2: $100C + 10B + A = 100A + 10B + C - 594$ We can simplify this equation by moving terms involving $A$ and $C$ to one side: $100C - C + 10B + A - 100A = -594$ This simplifies to: $99C - 99A = -594$ Dividing through by 99: $C - A = -6$ or equivalently, $C = A - 6$

Step 2: Substitute $C = A - 6$ into Condition 1

Since $A + B + C = 12$, we can substitute $C$ in terms of $A$: $A + B + (A - 6) = 12$ This simplifies to: $2A + B = 18$

Step 3: Solve for Possible Values of $A$, $B$, and $C$

Since $A$, $B$, and $C$ are digits (0–9), and $A$ is the hundreds place (thus $A \geq 1$), let's check feasible values for $A$:

1. If $A = 7$: $2(7) + B = 18$ $14 + B = 18 \Rightarrow B = 4$ Then, $C = A - 6 = 7 - 6 = 1$.

   This gives $A = 7$, $B = 4$, $C = 1$, and the number $ABC = 741$.

Verification

Let's verify that $ABC = 741$ satisfies both conditions:

1. Sum of Digits: $A + B + C = 7 + 4 + 1 = 12$.
2. Reversed Number Condition: The reversed number $CBA = 147$. $741 - 147 = 594$ Both conditions are satisfied.

Conclusion

The original number $ABC$ is: $\boxed{741}$

Would you like more details, or have any questions about the steps?

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1. How would the solution change if the difference between $ABC$ and $CBA$ was a different number?
2. Could there be multiple solutions if the conditions were slightly altered?
3. How does reversing the digits affect the value of the number mathematically?
4. What would happen if we changed the sum condition $A + B + C = 12$ to another number?
5. Can this type of problem be generalized to four-digit numbers?

Tip: When solving digit-based problems, expressing each digit’s place value as an algebraic expression can simplify complex conditions.