Existence check with H.C.F and L.C.M: Consider pairs of integers that have H.C.F = 16 and L.C.M = 136. What can we definitely say about such pairs?

Introduction / Context:This question asks whether any pair of integers can have the stated HCF and LCM. A necessary condition for two integers a and b is a * b = HCF * LCM. Additionally, the LCM must be a multiple of the HCF.

Given Data / Assumptions:

• HCF = 16
• LCM = 136
• Integers a, b > 0 assumed.

Concept / Approach:Write a = 16x and b = 16y with HCF(x, y) = 1. Then LCM(a, b) = 16xy. For LCM to be 136, we need 16xy = 136 ⇒ xy = 136/16 = 8.5, which is not an integer. Hence impossible.

Step-by-Step Solution:

Verification / Alternative check:Check divisibility: LCM must be a multiple of HCF. 136 is not divisible by 16 (136/16 = 8.5), so it cannot be an LCM when HCF is 16.

Why Other Options Are Wrong:If no pair exists, counts like “one” or “two” or “many” are impossible.

Common Pitfalls:Forgetting the requirement that LCM must be a multiple of HCF, or ignoring the integral nature of xy in the representation a = 16x, b = 16y.

Final Answer:No such pair exists