Two positive integers have a sum of 1215 and their highest common factor (HCF) is 81. How many unordered pairs of such numbers exist, and what are all the pairs?

Introduction / Context:This problem uses the relationship between a pair of integers, their highest common factor (HCF), and their sum. By factoring out the HCF, we reduce the question to counting coprime pairs that add to a fixed total. We then lift those back to the original numbers.

Given Data / Assumptions:

• Sum of the numbers = 1215
• HCF (greatest common divisor) = 81
• Numbers are positive integers
• We count unordered pairs (order does not matter)

Concept / Approach:Let the numbers be 81a and 81b where gcd(a, b) = 1. Then 81a + 81b = 1215 ⇒ a + b = 1215 / 81. Once we find a + b, we list all positive coprime pairs (a, b) with that sum and convert to the original numbers (81a, 81b). Each unordered coprime pair yields one valid pair of original numbers.

Step-by-Step Solution:1215 / 81 = 15 ⇒ a + b = 15, with gcd(a, b) = 1.List positive unordered pairs summing to 15: (1,14), (2,13), (3,12), (4,11), (5,10), (6,9), (7,8).Keep only coprime pairs: (1,14), (2,13), (4,11), (7,8). (Pairs with 3,12 and 6,9 and 5,10 are not coprime.)Multiply each by 81: (81,1134), (162,1053), (324,891), (567,648).Hence, there are 4 unordered pairs.

Verification / Alternative check:All listed pairs sum to 1215 and each pair has HCF 81, since gcd(81a, 81b) = 81*gcd(a, b) = 81 when gcd(a, b) = 1.

Why Other Options Are Wrong:Answers 1, 2, and 3 underestimate the number of coprime decompositions of 15; 5 overestimates by including at least one non-coprime pair.

Common Pitfalls:Forgetting to enforce gcd(a, b) = 1; double-counting ordered pairs as two; or missing one of the valid coprime pairs that sum to 15.

Final Answer:4 (pairs: (81,1134), (162,1053), (324,891), (567,648))