The product of the digits of a two-digit number is 18. If we add 63 to this number, the new number formed has its digits interchanged. What is the original two-digit number?

Introduction / Context:
This puzzle is similar in style to other digit problems: a two-digit number is described by conditions on the product of its digits and on how the number changes when a fixed value is added. Translating these conditions into equations involving the tens and units digits is a key skill in quantitative aptitude, especially in the topic of number puzzles and digit manipulations.

Given Data / Assumptions:

• Let the two-digit number have tens digit a and units digit b.
• The number is 10a + b.
• The product of the digits is a * b = 18.
• When 63 is added to the number, the digits get interchanged, resulting in 10b + a.
• We must find the original number 10a + b.

Concept / Approach:
The number with digits interchanged is 10b + a. Using the condition that adding 63 to the original gives this reversed number leads to a linear equation in a and b. Combined with the product condition a * b = 18, this forms a system of two equations. Solving this system reveals the specific digits. This approach is systematic and avoids random guessing, though we can later confirm with the options easily.

Step-by-Step Solution:
Step 1: Let the original two-digit number be 10a + b, where a is the tens digit and b is the units digit.Step 2: We are given that a * b = 18.Step 3: When we add 63 to the original number, the digits are interchanged. So 10a + b + 63 = 10b + a.Step 4: Rearrange the equation: 10a + b + 63 = 10b + a.Step 5: Bring all terms to one side: 10a - a + b - 10b + 63 = 0.Step 6: Simplify: 9a - 9b + 63 = 0.Step 7: Factor out 9: 9(a - b + 7) = 0, so a - b + 7 = 0.Step 8: Therefore, a = b - 7.Step 9: Use the product condition a * b = 18. Substitute a = b - 7, giving (b - 7) * b = 18.Step 10: Expand: b^2 - 7b - 18 = 0.Step 11: Solve the quadratic b^2 - 7b - 18 = 0. The discriminant is 49 + 72 = 121, whose square root is 11.Step 12: The roots are b = (7 ± 11) / 2, giving b = 9 or b = -2.Step 13: A digit must be between 0 and 9, so b = 9 is valid and b = -2 is invalid.Step 14: Then a = b - 7 = 9 - 7 = 2.Step 15: Thus, the original number is 10a + b = 10 * 2 + 9 = 29.

Verification / Alternative check:
Check the product of the digits: 2 * 9 = 18, matching the first condition. Now add 63 to the number: 29 + 63 = 92. The number 92 has digits 9 and 2, which are the reverse of the original digits 2 and 9. Therefore, both the product condition and the digit reversal condition hold, confirming that 29 is the correct original number.

Why Other Options Are Wrong:
Option 92: This is actually the number obtained after adding 63 to the original, not the original itself.
Option 36: Digits are 3 and 6, product is 18, but 36 + 63 = 99, which does not have digits 6 and 3.
Option 63: Digits are 6 and 3, product is 18, but 63 + 63 = 126, which is not the reverse 36.
Option 39: Digits are 3 and 9, product is 27, not 18.

Common Pitfalls:
Some students may directly test each option but forget to verify both conditions (product and reversal after addition). Others might mix up the digits in the equation and write 10b + a + 63 = 10a + b instead of the correct relation. Also, errors often occur in solving the quadratic equation if the discriminant or roots are computed incorrectly. Careful algebra and a quick double-check with the given conditions help avoid these mistakes.

Final Answer:
The original two-digit number is 29.