Factors
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Core Concept: Every composite integer is built from prime numbers, acting as the fundamental "chemical elements" of arithmetic.
Imagine packaging batches of macadamia nuts in Meru. If you have 12 nuts, you can pack them in equal bags without remainders: 1 bag of 12, 2 bags of 6, 3 bags of 4, 4 bags of 3, 6 bags of 2, or 12 bags of 1. Each complete pack size is a factor of 12.
1. Geometric Model of Factors
A factor pair \(a \times b = n\) represents a whole-number rectangle of area \(n\). If a number is prime (like 2, 3, 5, 7, 11), it can only form a single straight line rectangle \(1 \times p\).
2. The Fundamental Theorem of Arithmetic
Every integer \(n > 1\) can be uniquely expressed as a product of prime powers: \[ n = p_1^{a_1} \times p_2^{a_2} \times \dots \times p_k^{a_k} \] For example, \(60 = 2^2 \times 3^1 \times 5^1\). No other set of primes can multiply to give 60.
3. Visualising HCF and LCM via Prime Factors
- Highest Common Factor (HCF / GCD): The product of the lowest exponent of each shared prime factor (the intersection of their prime building blocks).
- Lowest Common Multiple (LCM): The product of the highest exponent of all prime factors appearing in either number (the union of their prime building blocks).
Interactive Prime Explorer: 12 vs 18
\(2^2 \times 3^1\)
\(2^1 \times 3^1 = 6\)
\(2^1 \times 3^2\)
\(2^2 \times 3^2 = 36\)
Key Formulas
1. Prime Factorisation Form
\[ n = p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k} \]Where \(p_1, p_2, \dots, p_k\) are distinct prime numbers and \(e_i \ge 1\) are positive integer exponents.
2. Total Number of Factors
\[ \tau(n) = (e_1 + 1)(e_2 + 1)\cdots(e_k + 1) \]Gives the exact count of all positive divisors of \(n\).
3. Highest Common Factor (HCF / GCD)
\[ \text{HCF}(a, b) = \prod p_i^{\min(e_i^{(a)},\, e_i^{(b)})} \]Select each common prime factor with its lowest exponent.
4. Lowest Common Multiple (LCM)
\[ \text{LCM}(a, b) = \prod p_i^{\max(e_i^{(a)},\, e_i^{(b)})} \]Select every prime factor present with its highest exponent.
5. Fundamental Product Identity
\[ \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \]Applies to any two positive integers \(a\) and \(b\).
Worked Examples
Example 1 (Easy): Finding HCF using Factor Trees
Problem: Find the Highest Common Factor (HCF) of 24 and 36.
- Step 1: Express 24 in prime factor form: \[ 24 = 2 \times 12 = 2 \times 2 \times 6 = 2^3 \times 3^1 \]
- Step 2: Express 36 in prime factor form: \[ 36 = 6 \times 6 = (2 \times 3) \times (2 \times 3) = 2^2 \times 3^2 \]
- Step 3: Take the minimum power of each common prime factor: \[ \text{HCF}(24, 36) = 2^{\min(3, 2)} \times 3^{\min(1, 2)} = 2^2 \times 3^1 = 4 \times 3 = 12 \]
Answer: \(\mathbf{12}\)
Example 2 (Medium): Calculating Total Divisors and LCM
Problem: Find the prime factorisation of 180, count its total number of factors, and determine \(\text{LCM}(180, 75)\).
- Step 1: Prime factorise 180: \[ 180 = 18 \times 10 = (2 \times 3^2) \times (2 \times 5) = 2^2 \times 3^2 \times 5^1 \]
- Step 2: Total number of factors = \((2+1)(2+1)(1+1) = 3 \times 3 \times 2 = 18\) factors.
- Step 3: Prime factorise 75: \[ 75 = 3 \times 25 = 3^1 \times 5^2 \]
- Step 4: Determine LCM using highest powers: \[ \text{LCM}(180, 75) = 2^2 \times 3^2 \times 5^2 = 4 \times 9 \times 25 = 900 \]
Answer: Factors of 180 = \(\mathbf{18}\); LCM = \(\mathbf{900}\)
Example 3 (Hard): Real-World Dual Cycle Synchronization
Problem: At a bus terminal in Nairobi, bus line A departs every 42 minutes, and bus line B departs every 70 minutes. If both leave simultaneously at 6:00 AM, after how many minutes will they next depart together?
- Step 1: Identify the requirement: The next simultaneous departure occurs at the Least Common Multiple (LCM) of 42 and 70 minutes.
- Step 2: Prime factorise both time intervals: \[ 42 = 2 \times 3 \times 7 = 2^1 \times 3^1 \times 7^1 \] \[ 70 = 2 \times 5 \times 7 = 2^1 \times 5^1 \times 7^1 \]
- Step 3: Calculate LCM by picking the highest power of every prime present: \[ \text{LCM}(42, 70) = 2^1 \times 3^1 \times 5^1 \times 7^1 = 2 \times 3 \times 5 \times 7 = 210 \text{ minutes} \]
- Step 4: Check using \(\text{HCF} \times \text{LCM} = a \times b\): \[ \text{HCF}(42, 70) = 2^1 \times 7^1 = 14 \] \[ 14 \times 210 = 2940 \quad \text{and} \quad 42 \times 70 = 2940 \quad \text{(Matches!)} \]
Answer: \(\mathbf{210}\) minutes (which is 3 hours 30 minutes later, at 9:30 AM).
Common Mistakes
Misconception 1: Confusing HCF with LCM
Misconception 2: Forgetting 1 as a Universal Factor
Misconception 3: Believing Primes Have Only 1 Factor
Real World
Practice