Let's denote the original two-digit number as $10x + y$, where $x$ is the tens digit and $y$ is the units digit.

Step 1: Formulating the First Equation

According to the problem, one of the digits is four times the other. We have two cases:

1. $x = 4y$
2. $y = 4x$

Let's handle both cases.

Step 2: Formulating the Second Equation

When the digits are interchanged, the new number becomes $10y + x$. The sum of the new number and the original number is given as 55. Therefore, the equation is: $(10x + y) + (10y + x) = 55$ Simplifying this equation: $11x + 11y = 55$ Dividing both sides by 11: $x + y = 5$

Step 3: Solving for Both Cases

Case 1: $x = 4y$

Substitute $x = 4y$ into the equation $x + y = 5$: $4y + y = 5$ $5y = 5$ $y = 1$ Now, substitute $y = 1$ back into $x = 4y$: $x = 4 \times 1 = 4$ Thus, the original number is $10x + y = 10 \times 4 + 1 = 41$.

Case 2: $y = 4x$

Substitute $y = 4x$ into the equation $x + y = 5$: $x + 4x = 5$ $5x = 5$ $x = 1$ Now, substitute $x = 1$ back into $y = 4x$: $y = 4 \times 1 = 4$ Thus, the original number is $10x + y = 10 \times 1 + 4 = 14$.

Step 4: Conclusion

There are two possible original numbers: 41 and 14. Both satisfy the conditions of the problem.

Would you like further clarification on any step?

Here are 5 related questions:

1. How would the problem change if the sum of the numbers was 66 instead of 55?
2. Can we apply the same approach to a three-digit number with similar conditions?
3. What if the digits differ by 3 instead of one being a multiple of the other?
4. How can this method be adapted to non-decimal (e.g., binary or hexadecimal) systems?
5. What other conditions could be added to further limit the possible original numbers?

Tip: When faced with digit-related problems, try setting up equations based on place values (e.g., tens and units digits) for clarity.