M and N are two distinct natural numbers. Their HCF and LCM are K and L…

For any two natural numbers M and N with HCF K and LCM L, two identities always hold: (1) K × L = M × N — the product of the HCF and LCM equals the product of the two numbers; and (2) K always divides L — writing M = K·m and N = K·n with gcd(m, n) = 1, we get L = K·m·n.

1. Since M and N are distinct, m and n cannot both be 1 (that would force M = N = K). Since gcd(m, n) = 1, at least one of m, n exceeds 1, so m × n ≥ 2.

2. So L = K × m × n ≥ 2K — meaning L is always strictly greater than K whenever M ≠ N.

3. Check K × L = A: this needs A = M × N, which is always a natural number — this relation always holds for a suitable A.

4. Check K × A = L: this needs A = L ÷ K = m × n, which is exactly a whole number — this relation always holds for a suitable A.

5. Check L × A = K: this needs A = K ÷ L. Since L is strictly greater than K, K ÷ L is a proper fraction strictly between 0 and 1, which can never equal a natural number — no A satisfies this relation.

Verify with M = 4, N = 6: K = 2, L = 12. K × L = A gives A = 24 = M × N (holds). K × A = L gives A = 6 = m × n (holds, since m = 2, n = 3). L × A = K gives A = 2/12 = 1/6, not a natural number — confirming this relation is impossible.

Hence the relation L × A = K can never hold for a natural number A, making it the one that is not possible.