Algebraic Relations of HCF LCM for UPSC CSAT

1. Theoretical Foundation & Mathematical Underpinnings

Formal Definition & Scope

For a, b ∈ ℤ⁺ where a, b > 1:

• HCF (Highest Common Factor) / gcd(a, b) = H: The greatest integer H such that H | a and H | b. It contains only the common prime factors at their minimum exponent.
• LCM (Least Common Multiple) / lcm(a, b) = L: The smallest integer L such that a | L and b | L. It contains all prime factors of a and b at their maximum exponent.

Domain Constraint: By definition,

H | a,   H | b,   a | L,   b | L

and crucially,

H | L

If L mod H ≠ 0, the dataset is impossible — no such pair (a, b) exists.

The “Why” Behind The Math

Theorem 1: The Co-Prime Decomposition

Any two numbers can be decomposed with respect to their HCF.

Let H = gcd(a, b). Then there exist unique x, y ∈ ℤ⁺ such that:

a = Hx,   b = Hy   where   gcd(x, y) = 1

Why must gcd(x, y) = 1? — Proof by Contradiction:

Suppose gcd(x, y) = d > 1. Then x = dx′ and y = dy′ for some x′, y′ ∈ ℤ⁺.

Substituting these back:

a = H(dx′) = (Hd)x′
b = H(dy′) = (Hd)y′

This implies Hd is a common divisor of a and b. Since d > 1, we have Hd > H, which contradicts the premise that H is the Highest Common Factor. Hence d must equal 1, proving (x, y) is strictly co-prime.

Corollary:

^a/_b = ^Hx/_Hy = ^x/_y

The ratio in lowest terms is the co-prime pair (x, y).

Theorem 2: The Two-Variable Product Invariant

For exactly two numbers, the foundational identity holds:

H × L = a × b

Proof 1 — Via Co-Prime Structure (Intuitive):

From Theorem 1, the least common multiple must contain the common base H alongside all distinct non-common factors x and y:

L = Hxy

Multiplying by H:

H × L = H × (Hxy)
H × L = H²xy

Expanding the product of the original numbers:

a × b = (Hx) × (Hy)
a × b = H²xy

Thus, H × L = a × b.

Proof 2 — Via Prime Factorization (Rigorous):

Let the prime factorizations of a and b be:

a = ∏ p_i^α_i
b = ∏ p_i^β_i

By definition:

H = ∏ p_i^{min(α_i, β_i)}
L = ∏ p_i^{max(α_i, β_i)}

Since for any two real numbers min(α_i, β_i) + max(α_i, β_i) = α_i + β_i:

H × L = ∏ p_i^{min(α_i, β_i) + max(α_i, β_i)}
H × L = ∏ p_i^{α_i + β_i}
H × L = (∏ p_i^α_i) (∏ p_i^β_i)
H × L = a × b

Worked Verification:
Let a = 12 and b = 15.
• H = gcd(12, 15) = 3
• 12 = 3 × 4 ⇒ x = 4
• 15 = 3 × 5 ⇒ y = 5
• gcd(4, 5) = 1 is satisfied.
• L = Hxy = 3 × 4 × 5 = 60

Verifying the invariant:
H × L = 3 × 60 = 180
a × b = 12 × 15 = 180

Key Theorems & Properties

• Divisibility Invariant: H | L always. If ^L/_H ∉ ℤ⁺, no valid integer pair (a, b) exists.
  Example: H = 14, L = 200 ⇒ ²⁰⁰/₁₄ ∉ ℤ ⇒ Zero valid pairs exist.
• Co-Prime Quotient Invariant: ^L/_H = xy where gcd(x, y) = 1. This quotient serves as the core computational engine for all pair-counting problems.
• Exponent Complementarity Rule: For two numbers, min(α_i, β_i) + max(α_i, β_i) = α_i + β_i. This confirms why the product rule strictly holds for n = 2.
• Structural Breakdown for n ≥ 3: For three numbers a, b, c:
  H = ∏ p_i^{min(α_i, β_i, γ_i)}
  L = ∏ p_i^{max(α_i, β_i, γ_i)}
  Because min(α_i, β_i, γ_i) + max(α_i, β_i, γ_i) ≠ α_i + β_i + γ_i, the identity fails:
  H × L ≠ a × b × c
• Ratio Invariance: If ^a/_b = ^p/_q in lowest terms, then a = Hp and b = Hq.

2. Comparative Matrix & Conceptual Distinctions

Table 1: HCF vs. LCM — Structural Duality

Table 2: Comparative Analysis — Two Numbers vs. Three Numbers

3. High-Yield Data Anchors & Memorization Benchmarks

Anchor 1: The Divisibility Filter

Before executing full algebraic calculations, evaluate whether L mod H = 0. If the remainder is non-zero, the configuration is invalid, yielding zero valid pairs.

Anchor 2: The Co-Prime Distribution Law

Let N = ^L/_H = xy with the strict constraint gcd(x, y) = 1.
Let k be the number of distinct prime factors of N.

Formula Benchmarks:
• Number of unordered pairs {x, y} (and thus {a, b}) = 2^k-1
• Number of ordered pairs (x, y) = 2^k
(For N = 1 where L = H, there is exactly 1 pair: a = b = H. The formula 2^k-1 applies for all k ≥ 1.)

• Example 1:
  H = 12, L = 924 ⇒ N = ⁹²⁴/₁₂ = 77 = 7¹ × 11¹.
  Here k = 2, so number of pairs = 2^2-1 = 2.
• Example 2:
  H = 15, L = 300 ⇒ N = ³⁰⁰/₁₅ = 20 = 2² × 5¹.
  Here k = 2, so number of pairs = 2^2-1 = 2.
  The factor pairs of 20 are (1, 20) and (4, 5). The pair (2, 10) is rejected because gcd(2, 10) = 2 ≠ 1.

Anchor 3: Common Non-Co-Prime Quotient Traps

• N = 12: Divisor factor pairs are (1, 12), (2, 6), and (3, 4).
  The pair (2, 6) is invalid because gcd(2, 6) = 2.
  ⇒ Exactly 2 valid pairs remain: (1, 12) and (3, 4).
• N = 36: Divisor factor pairs are (1, 36), (2, 18), (3, 12), (4, 9), and (6, 6).
  Pairs (2, 18), (3, 12), and (6, 6) share common divisors.
  ⇒ Exactly 2 valid pairs remain: (1, 36) and (4, 9).

4. Standard Algorithmic Protocols

Protocol A: Systematically Finding Pairs Given H and L

Step-by-Step Procedure

1. Divisibility Verification: Check if L mod H = 0.
   • If No: Terminate. Zero valid pairs exist.
   • If Yes: Proceed to Step 2.
2. Calculate Quotient: Compute N = ^L/_H.
3. Prime Factorization: Express N = p₁^a₁ · p₂^a₂ ··· p_k^a_k and identify k (the count of distinct prime bases).
4. Count Configurations:
   Unordered pairs = 2^k-1,   Ordered pairs = 2^k
5. Reconstruct Numbers (if required):
   • Enumerate all factor pairs (d, ^N/_d) such that gcd(d, ^N/_d) = 1.
   • Multiply each component by H to get (Hd, H · (^N/_d)).

Numerical Execution

Find all pairs of natural numbers with H = 18 and L = 540.

• Step 1: 540 mod 18 = 0 (Valid).
• Step 2: N = ⁵⁴⁰/₁₈ = 30.
• Step 3: 30 = 2¹ × 3¹ × 5¹ ⇒ k = 3.
• Step 4: Total unordered pairs = 2^3-1 = 4.
• Step 5: The co-prime pairs (x, y) summing to product 30 are:
  (1, 30),   (2, 15),   (3, 10),   (5, 6)
  Multiplying by H = 18 yields the actual number pairs:
  (18, 540),   (36, 270),   (54, 180),   (90, 108)

Protocol B: Prime Exponent Matching for Three Numbers

For three numbers a, b, c, avoid the false product formula H × L = a × b × c. Use prime exponent bounding:

1. Factorize all known terms:

   a = ∏ p_i^α_i,   b = ∏ p_i^β_i
   H = ∏ p_i^H_i,   L = ∏ p_i^L_i

2. Apply boundary conditions per prime:

   min(α_i, β_i, γ_i) = H_i
   max(α_i, β_i, γ_i) = L_i

3. Determine constraints on c = ∏ p_i^γ_i:
   • If neither α_i nor β_i reaches L_i, then γ_i must equal L_i.
   • If neither α_i nor β_i drops to H_i, then γ_i must equal H_i.
   • Otherwise, γ_i can be any integer such that H_i ≤ γ_i ≤ L_i.

5. Standard Question Typologies & Analytical Solutions

Type A: Parity & Multi-Variable Constraints

Core Principle: When a = Hx and b = Hy, the algebraic parity of a and b is directly governed by H. If H is even, both a and b must be even.

Question: The HCF of two natural numbers is 12 and their LCM is 360. If one number is even and the sum of the numbers is odd, which of the following must be true?
(a) Both x and y are odd
(b) One of x, y is even, the other is odd
(c) Both x and y are even
(d) No such pair exists

Analytical Solution:

Step 1 — Structural Formulation:
Let the numbers be a = 12x and b = 12y, where gcd(x, y) = 1 and:
xy = ^L/_H = ³⁶⁰/₁₂ = 30

Step 2 — Parity Evaluation:
The sum of the numbers is:
a + b = 12x + 12y = 12(x + y) = 2 · [6(x + y)]
Since 12 is even, 12(x + y) is strictly even for all integers x, y. Therefore, an odd sum is mathematically impossible.

Step 3 — Conclusion:
Although factor pairs for xy = 30 exist, none can satisfy the condition that a + b is odd.

Correct Option: (d)

Quick Check: If H is even, then both a and b are even, making their sum and difference unconditionally even.

Type B: Multi-Statement Analytical Questions

Core Principle: Validate statements independently by calculating N = ^L/_H, applying co-prime filtering, and summing real pairs.

Question: Consider two natural numbers whose HCF is 15 and LCM is 300.
Statement I: There are exactly two possible pairs of numbers.
Statement II: The sum of the numbers in one valid pair is 135.
Which of the statements given above is/are correct?
(a) I only
(b) II only
(c) Both I and II
(d) Neither I nor II

Analytical Solution:

Step 1 — Quotient and Factorization:
N = ^L/_H = ³⁰⁰/₁₅ = 20 = 2² × 5¹

Step 2 — Co-Prime Factor Pair Extraction:
The factor pairs of 20 are:
(1, 20) ⇒ gcd(1, 20) = 1   (Valid)
(2, 10) ⇒ gcd(2, 10) = 2 ≠ 1   (Invalid)
(4, 5) ⇒ gcd(4, 5) = 1   (Valid)
There are exactly 2 valid co-prime pairs. Statement I is correct.

Step 3 — Reconstructing the Values:
• Pair 1: (15 × 1, 15 × 20) = (15, 300) ⇒ Sum = 315
• Pair 2: (15 × 4, 15 × 5) = (60, 75) ⇒ Sum = 135
Statement II is correct.

Correct Option: (c)

Type C: Pair Counting & Existence Archetypes

Core Principle: The number of unordered co-prime pairs depends directly on 2^k-1, where k is the count of distinct prime factors of ^L/_H.

Question: The HCF and LCM of two natural numbers are 12 and 924 respectively. How many such distinct pairs of numbers are possible?
(a) 1
(b) 2
(c) 3
(d) 4

Analytical Solution:

Step 1 — Compute Quotient:
N = ^L/_H = ⁹²⁴/₁₂ = 77

Step 2 — Distinct Prime Counting:
77 = 7¹ × 11¹ ⇒ k = 2

Step 3 — Calculate Unordered Pairs:
Number of distinct pairs = 2^k-1 = 2^2-1 = 2
The valid pairs are (x, y) ∈ {(1, 77), (7, 11)}, yielding (a, b) ∈ {(12, 924), (84, 132)}.

Correct Option: (b)

Type D: Ratio Invariance & Multiplier Logic

Core Principle: A simplified ratio a : b = p : q directly defines the co-prime pair x = p and y = q.

Question: Two numbers a and b have an HCF of 6. Their ratio in simplest form is 4 : 5. What is their LCM?
(a) 60
(b) 120
(c) 180
(d) 240

Analytical Solution:

Step 1 — Base Formulation:
Since 4 : 5 is irreducible, gcd(4, 5) = 1.
Thus, x = 4 and y = 5.

Step 2 — Calculate LCM:
L = H × x × y = 6 × 4 × 5 = 120
Alternatively, find the numbers first:
a = 6 × 4 = 24,   b = 6 × 5 = 30
L = ^{(a × b)}/_H = ^{(24 × 30)}/₆ = 120

Correct Option: (b)

Type E: Factor Splitting & Structural Validity

Core Principle: Prime powers in N = ^L/_H cannot be split across both x and y; each prime power block must reside entirely within one term.

Question: The HCF of two numbers is 18 and their LCM is 324. Which of the following cannot be the values of the two numbers?
(a) 18 and 324
(b) 36 and 162
(c) 54 and 108
(d) 72 and 81

Analytical Solution:

Step 1 — Factor Constraints:
N = ^L/_H = ³²⁴/₁₈ = 18 = 2¹ × 3²
To preserve gcd(x, y) = 1:

• The factor 2¹ must go entirely to x or y.
• The factor 3² must go entirely to x or y.

Valid (x, y) pairs are strictly (1, 18) and (2, 9).
Valid number pairs are (18, 324) and (36, 162).

Step 2 — Option Audit:

• Option (a): (18, 324) ⇒ (x, y) = (1, 18) with gcd(1, 18) = 1 (Valid).
• Option (b): (36, 162) ⇒ (x, y) = (2, 9) with gcd(2, 9) = 1 (Valid).
• Option (c): (54, 108) ⇒ (x, y) = (3, 6) with gcd(3, 6) = 3 ≠ 1 (Invalid — prime 3 is split).
• Option (d): 81 is not divisible by H = 18 (Fails basic divisibility H | b).

Both (c) and (d) are mathematically invalid; option (c) represents the classic co-prime factorization trap.

Intended Option: (c)

Type F: The Three-Variable Constraint Model

Core Principle: For three variables, prime factors missing from two numbers must be fully supplied by the third number to achieve the target LCM.

Question: The HCF of three numbers a, b, c is 4 and their LCM is 240. If a = 12 and b = 16, which of the following could be the value of c?
(a) 5
(b) 10
(c) 15
(d) 20

Analytical Solution:

Step 1 — Prime Factorization:
a = 12 = 2² × 3¹ × 5⁰
b = 16 = 2⁴ × 3⁰ × 5⁰
H = 4 = 2² × 3⁰ × 5⁰
L = 240 = 2⁴ × 3¹ × 5¹

Step 2 — Prime Exponent Constraints on c = 2^γ₁ × 3^γ₂ × 5^γ₃:
• Prime 5: max(0, 0, γ₃) = 1 ⇒ γ₃ = 1 (Factor of 5 is mandatory).
• Prime 2: min(2, 4, γ₁) = 2 and max(2, 4, γ₁) = 4 ⇒ 2 ≤ γ₁ ≤ 4 (Must be a multiple of 2² = 4).
• Prime 3: min(1, 0, γ₂) = 0 and max(1, 0, γ₂) = 1 ⇒ 0 ≤ γ₂ ≤ 1.

Step 3 — Minimum Feasible Value:
c_min = 2² × 3⁰ × 5¹ = 20

Checking options:

• (a) 5: Not divisible by H = 4.
• (b) 10: Divisible by 2, but not by 4.
• (c) 15: Not divisible by 4.
• (d) 20 = 2² × 5¹: Satisfies all constraints.

Correct Option: (d)

6. The Examiner Trap Matrix

7. Golden Rules & Instant Problem Triggers