To solve the problem, let's denote:

1. The two-digit number as $10x + y$, where:
   • $x$ is the tens digit
   • $y$ is the units digit.

Given Conditions:

1. The sum of the digits is 12: $x + y = 12$

2. The number obtained by interchanging the digits is $10y + x$, and it exceeds the original number by 18: $10y + x = 10x + y + 18$

Solving the Equations:

From the second equation:

$10y + x = 10x + y + 18$ Rearranging terms: $10y - y = 10x - x + 18$ $9y = 9x + 18$ $y = x + 2$

Substitute $y = x + 2$ into the first equation:

$x + (x + 2) = 12$ $2x + 2 = 12$ $2x = 10$ $x = 5$

Find $y$:

$y = x + 2 = 5 + 2 = 7$

The Original Number:

$10x + y = 10(5) + 7 = 57$

Final Answer:

The original number is 57.

Would you like further clarification? Here are 5 related questions to explore:

1. What if the sum of the digits was different (e.g., 15)? How would that affect the solution?
2. How would you approach this problem if the digits were reversed and subtracted instead of added?
3. Can you verify the solution by substituting the number into the original conditions?
4. How can this method be generalized for any two-digit number problem with similar constraints?
5. What are some real-world applications of such digit manipulation problems?

Tip: Always check your solution by substituting back into the original equations to confirm accuracy.