Advanced HCF-LCM | HCF and LCM Aptitude Practice | Kuriloo

Fundamental Principles

Coprime Pair Constraint System

Since any two numbers can be written as $a = xH$ and $b = yH$, then their product equation expands to: $(xH)(yH) = H \cdot LCM$, which simplifies to: $x \cdot y = LCM / H$.

Essential Formulation Tips

When finding valid number pairs, the values for x and y must be integers and they must be coprime ($HCF(x, y) = 1$).
The value of the LCM must be perfectly divisible by the HCF. If it isn't, no valid numbers can exist that satisfy those conditions.

Shortcut Execution Techniques

Pair Identification Flow: Divide the LCM by the HCF to find the target product ($x \cdot y$). Then, list out all possible factor pairs for that product and filter out any pairs that share a common factor.

Contextual Inquiries (FAQs)

Q: Can any random pair of HCF and LCM values form a valid set of numbers?

A: No. A combination is only valid if the LCM is perfectly divisible by the HCF without leaving a remainder.

Example Breakdown: Calculating Valid Coprime Number Pairs

Advanced constraint mapping question.

Problem QueryFind the total number of unique matching pairs of numbers that can have an HCF of 15 and an LCM of 90.

Step-By-Step Solution Path

Verify divisibility: 90 / 15 = 6 (perfect division, so valid combinations exist).

Set up the factor equation: $x \cdot y = LCM / HCF \rightarrow x \cdot y = 90 / 15 = 6$.

Find all positive integer pairs that multiply to 6: (1, 6) and (2, 3).

Check for coprime status: Both pairs are coprime since HCF(1,6) = 1 and HCF(2,3) = 1.

Count the final results: There are exactly 2 valid pairs.

Analytical Hint: Divide the LCM by the HCF, then find how many coprime factor pairs can multiply to equal that result.

Advanced Pair Constraints

Practice finding and verifying valid number pairs under strict factor constraints.

Q1. The HCF of two numbers is 12, and their sum is 72. How many unique pairs of numbers can satisfy these conditions?