The product of the digits of a two-digit number is 72. If we add 9 to the number, the resulting number has its digits interchanged. What is this original two-digit number?

Introduction / Context:
This question is a classic digit-based puzzle often found in quantitative aptitude tests. It involves a two-digit number with a constraint on the product of its digits and an additional condition about what happens when a fixed value is added to the number. The task is to translate these verbal conditions into equations involving the digits and then solve them systematically.

Given Data / Assumptions:

• Let the two-digit number have tens digit a and units digit b.
• The number itself can be written as 10a + b.
• The product of its digits is a * b = 72.
• When 9 is added to the number, we get a new number whose digits are interchanged, that is 10b + a.
• We must find the original two-digit number.

Concept / Approach:
We use place value to express the original and the reversed numbers. The original is 10a + b, while the reversed number is 10b + a. Using the condition that adding 9 to the original number gives the reversed number, we can write an equation involving a and b. Together with the product condition a * b = 72, this forms a solvable system. Solving these equations will give the digits a and b and thus the original number.

Step-by-Step Solution:
Step 1: Let the tens digit be a and the units digit be b. Then the number is 10a + b.Step 2: The product of the digits is a * b = 72.Step 3: When 9 is added to the number, we obtain the reversed-digit number: 10a + b + 9 = 10b + a.Step 4: Rearrange the equation: 10a + b + 9 = 10b + a.Step 5: Bring like terms together: 10a - a + b - 10b + 9 = 0, which simplifies to 9a - 9b + 9 = 0.Step 6: Factor out 9: 9(a - b + 1) = 0, so a - b + 1 = 0.Step 7: Therefore, a = b - 1.Step 8: Use the product condition a * b = 72. Substitute a = b - 1 to get (b - 1) * b = 72.Step 9: Expand: b^2 - b - 72 = 0.Step 10: Solve the quadratic equation b^2 - b - 72 = 0. The discriminant is 1 + 288 = 289, whose square root is 17.Step 11: The roots are b = (1 ± 17) / 2. This gives b = 9 or b = -8.Step 12: A digit must be non-negative and at most 9, so b = 9 is valid and b = -8 is discarded.Step 13: Then a = b - 1 = 9 - 1 = 8.Step 14: Hence the original two-digit number is 10a + b = 10 * 8 + 9 = 89.

Verification / Alternative check:
Check the product of digits: 8 * 9 = 72, which matches the first condition. Now add 9 to the number: 89 + 9 = 98. The new number 98 has digits 9 and 8, which are exactly the reverse of the original digits 8 and 9. Thus, both conditions are satisfied, confirming that 89 is correct.

Why Other Options Are Wrong:
Option 98: Product of digits is 9 * 8 = 72, but if we add 9 to 98, we get 107, which does not have digits reversed.
Option 78: Product of digits is 7 * 8 = 56, not 72.
Option 87: Product of digits is 8 * 7 = 56, again not 72.
Option 69: Product is 6 * 9 = 54, and adding 9 gives 78, whose digits are not the reverse of 69.

Common Pitfalls:
Some students may attempt to test each option without constructing equations, which can be slower. Others might misinterpret the reversal condition and write 10a + b - 9 = 10b + a or confuse the roles of a and b. Another common error is making algebraic mistakes when simplifying the equations. Setting up the equations carefully and solving step by step avoids these issues.

Final Answer:
The original two-digit number is 89.