The sum of two numbers is 1056 and their HCF is 66. How many unordered pairs of such numbers exist?

Introduction:
Represent numbers with a common HCF using coprime multipliers. Then convert the sum condition into a constraint on those multipliers, and count the coprime solutions accordingly.

Given Data / Assumptions:

• Let the numbers be 66a and 66b
• HCF(66a, 66b) = 66 implies HCF(a, b) = 1
• 66a + 66b = 1056

Concept / Approach:
From the sum we get a + b = 1056 / 66 = 16. Count coprime positive pairs (a, b) with a + b = 16. For unordered pairs, count each pair once with a < b.

Step-by-Step Solution:

Verification / Alternative check:
These correspond to numbers (66, 990), (198, 858), (330, 726), and (462, 594), each with HCF 66 and sum 1056.

Why Other Options Are Wrong:
2 and 3 undercount; 6 and 8 double count ordered pairs or include noncoprime pairs.

Common Pitfalls:
Counting ordered pairs separately or including pairs like (2, 14) where the multipliers are not coprime.

Final Answer:
4