LCM feasibility given a fixed HCF If the HCF (gcd) of two positive integers is 26, which of the following cannot be their LCM?

Introduction / Context:
For two positive integers a and b, the fundamental identity is: a * b = HCF(a, b) * LCM(a, b). If HCF = 26, then both a and b are multiples of 26, say a = 26m and b = 26n with gcd(m, n) = 1. Consequently, LCM = 26 * m * n.

Given Data / Assumptions:

• HCF = 26.
• a = 26m, b = 26n, gcd(m, n) = 1.
• Therefore LCM must be 26 times an integer product of two coprime integers.

Concept / Approach:
Check each candidate whether it is a multiple of 26. If it is not divisible by 26, it cannot be the LCM. If it is divisible by 26, verify that the quotient can be written as a product of two coprime integers (always possible with 1 * k).

Step-by-Step Solution:
78 = 26 * 3 ⇒ possible (m, n) = (1, 3).104 = 26 * 4 ⇒ possible (m, n) = (1, 4).144 ÷ 26 is not an integer ⇒ impossible.234 = 26 * 9 ⇒ possible (m, n) = (1, 9).260 = 26 * 10 ⇒ possible (m, n) = (2, 5) with gcd 1.

Verification / Alternative check:
Because HCF divides both numbers, it must also divide their LCM. Any candidate LCM not divisible by 26 is invalid immediately.

Why Other Options Are Wrong:
All other options are multiples of 26 and are feasible.

Common Pitfalls:
Forgetting the divisibility requirement of LCM by the HCF, or believing m and n must both exceed 1; m = 1 is allowed.

Final Answer:
144