How many pairs of positive integers (x, y) with x ≤ y satisfy HCF(x, y) +…
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How many pairs of positive integers (x, y) with x ≤ y satisfy HCF(x, y) + LCM(x, y) = 91?
Answer: B. 8 — ConceptIf h = HCF(x, y), write x = ha and y = hb, where HCF(a, b) = 1. The corresponding LCM is hab. Thus an HCF–LCM sum can be converted into h(1 + ab). For…
- A.
10
- B.
8
- C.
6
- D.
7
Show answer & explanation
Correct answer: B
Concept
If h = HCF(x, y), write x = ha and y = hb, where HCF(a, b) = 1. The corresponding LCM is hab.
Thus an HCF–LCM sum can be converted into h(1 + ab). For an unordered count, the restriction x ≤ y is equivalent to a ≤ b, so a reversed pair is not counted again.
Application
Substitute HCF(x, y) = h and LCM(x, y) = hab into the given equation: h + hab = 91, so h(1 + ab) = 91.
Because h is a positive divisor of 91, test h ∈ {1, 7, 13, 91}. The case h = 91 gives ab = 0, which is impossible for positive a and b.
For each remaining h, solve ab = 91/h − 1 and retain only factor pairs with a ≤ b and HCF(a, b) = 1.
h | ab | Coprime factor pairs (a, b) | Resulting pairs (x, y) | Count |
|---|---|---|---|---|
1 | 90 | (1, 90), (2, 45), (5, 18), (9, 10) | (1, 90), (2, 45), (5, 18), (9, 10) | 4 |
7 | 12 | (1, 12), (3, 4) | (7, 84), (21, 28) | 2 |
13 | 6 | (1, 6), (2, 3) | (13, 78), (26, 39) | 2 |
Cross-check
The cases are exhaustive because every possible h must divide 91. For example, HCF(21, 28) = 7 and LCM(21, 28) = 84, whose sum is 91; the same construction verifies every listed pair.
Result
The total number of pairs is 4 + 2 + 2 = 8.