Let the two-digit number be represented as $10a + b$, where:

• $a$ is the tens digit, and
• $b$ is the ones digit.

Step 1: Set up the equations based on the given conditions.

• The sum of the digits is 8, so: $a + b = 8 \quad \text{(Equation 1)}$
• The number obtained by interchanging the digits exceeds the original number by 18. The number formed by interchanging the digits is $10b + a$. Hence, we can write: $10b + a = 10a + b + 18 \quad \text{(Equation 2)}$

Step 2: Simplify Equation 2.

Rearranging Equation 2: $10b + a - 10a - b = 18$ $9b - 9a = 18$ $b - a = 2 \quad \text{(Equation 3)}$

Step 3: Solve the system of equations.

Now, we have the system of two equations:

1. $a + b = 8$
2. $b - a = 2$

Solve for $b$ from Equation 3:

$b = a + 2$

Substitute $b = a + 2$ into Equation 1:

$a + (a + 2) = 8$ $2a + 2 = 8$ $2a = 6$ $a = 3$

Substitute $a = 3$ into $b = a + 2$:

$b = 3 + 2 = 5$

Step 4: Find the original number.

The original number is $10a + b = 10(3) + 5 = 35$.

Thus, the original number is 35.

Verification:

• The sum of the digits: $3 + 5 = 8$, which is correct.
• The number obtained by interchanging the digits is $53$, and $53 - 35 = 18$, which is also correct.

Therefore, the original number is 35.

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Would you like more details or have any questions?

Here are 5 related questions you can explore:

1. How would the solution change if the sum of the digits was different?
2. What happens if the number obtained by interchanging the digits exceeds the original number by a different value?
3. Can this method be used for three-digit numbers or larger?
4. What is the general approach to solving problems involving digit manipulation?
5. How would you solve if the difference between the interchanged number and the original number was less than 18?

Tip: When solving these kinds of problems, always start by assigning variables to the digits of the number, then translate the conditions into equations.