Let the original two-digit number be represented as:

$10a + b$ where:

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

We are given two pieces of information:

1. The sum of the digits is 8, so: $a + b = 8$
2. The number obtained by interchanging the digits exceeds the original number by 18. The number formed by interchanging the digits is $10b + a$, and the difference between this new number and the original number is 18: $(10b + a) - (10a + b) = 18$ Simplifying the second equation: $10b + a - 10a - b = 18$ $9b - 9a = 18$ $b - a = 2$

Now we have a system of two equations:

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

Solving the system:

From the second equation, solve for $b$: $b = a + 2$ Substitute this into the first equation: $a + (a + 2) = 8$ $2a + 2 = 8$ $2a = 6$ $a = 3$

Now substitute $a = 3$ into $b = a + 2$: $b = 3 + 2 = 5$

Thus, the original number is: $10a + b = 10(3) + 5 = 35$

Verification:

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

So, the original number is 35.

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Would you like further details or clarification on the solution?

Here are 5 related questions for deeper exploration:

1. What would happen if the sum of the digits were different, say 7 instead of 8?
2. How can this method be applied to a three-digit number?
3. How would you solve this type of problem using algebraic methods involving quadratic equations?
4. What happens if the difference between the original and interchanged number is a different value, such as 10 or 20?
5. How do we generalize this method for any number of digits?

Tip: When solving for digit-based problems, always express the number in terms of its individual digits (tens, ones, etc.) and set up equations based on the conditions provided.