A and B are positive integers with A, B ≤ 15 and A + B + AB = 65. What is the difference between A and B?

Introduction / Context:
This problem is a neat application of algebraic manipulation to transform a seemingly complicated expression into a factorised form. It appears often in aptitude exams under algebra and number system topics, using a classic trick A + B + AB = constant.

Given Data / Assumptions:

• A and B are positive integers.
• A, B ≤ 15.
• A + B + AB = 65.
• We must find the absolute difference between A and B.

Concept / Approach:
A key identity is: A + B + AB + 1 = (A + 1)(B + 1). By adding 1 to both sides of the given equation, we obtain a factorised form that allows us to find integers A and B as factors of 66. Then we can enforce the condition A, B ≤ 15 and compute the difference between the valid pair.

Step-by-Step Solution:
Given A + B + AB = 65. Add 1 to both sides: A + B + AB + 1 = 66. Recognise the factorisation (A + 1)(B + 1) = 66. Now list factor pairs of 66: (1, 66), (2, 33), (3, 22), (6, 11) and their reverses. So possible pairs for (A + 1, B + 1) are (1, 66), (2, 33), (3, 22), (6, 11), (11, 6), (22, 3), (33, 2), (66, 1). A and B must be positive and at most 15. Thus A + 1 and B + 1 must be at most 16. The only factor pair within this limit is (6, 11), giving A + 1 = 6, B + 1 = 11 or A + 1 = 11, B + 1 = 6. Hence (A, B) is either (5, 10) or (10, 5). The difference |A - B| = |5 - 10| = 5.

Verification / Alternative check:
Check with A = 5, B = 10: A + B + AB = 5 + 10 + 5*10 = 15 + 50 = 65, which satisfies the condition. Similarly, A = 10 and B = 5 gives the same sum, so the pair is valid and unique under the given constraints.

Why Other Options Are Wrong:
Differences 3, 4, 6 and 2 correspond to other factor pairs, but those factor pairs either do not multiply to 66 or lead to A or B values greater than 15 or non positive, which are not allowed.

Common Pitfalls:
One frequent mistake is to treat A + B + AB as (A + B)^2, which is incorrect. Another common error is to forget to add 1 before factorising. Recognising the standard transformation A + B + AB = (A + 1)(B + 1) - 1 is a powerful trick in many exam problems of this style.

Final Answer:
The required difference between A and B is 5.