2.4 Number Operations & Properties

Order of Operations & Number Classifications

Core Concept: Performing multi-step mathematical calculations requires following a standardised hierarchy of operations (BODMAS / PEMDAS) to ensure consistent and accurate results.

1. The Order of Operations (BODMAS / PEMDAS Rules)

When an expression involves multiple operations, solve them strictly according to operational priority:

The Operational Hierarchy Table

Crucial Operational Warning: Left-to-Right Rule

Division does NOT automatically precede multiplication, nor does addition automatically precede subtraction. When operations of equal priority appear together, evaluate them strictly from left to right.

• Example (Common Trap): $24 \div 6 \times 2$.
  • Correct: Work left to right: $24 \div 6 = 4$, then $4 \times 2 = \mathbf{8}$.
  • Incorrect: Multiplying first: $6 \times 2 = 12$, then $24 \div 12 = 2$ (WRONG!).

Detailed Worked Example

Problem: Evaluate $36 - 4 \times (8 - 3) + 2^3 \div 4$.

Step-by-Step Solution:

1. Brackets: Evaluate $(8 - 3) = 5$. $$\text{Expression: } 36 - 4 \times 5 + 2^3 \div 4$$
2. Orders / Exponents: Evaluate $2^3 = 8$. $$\text{Expression: } 36 - 4 \times 5 + 8 \div 4$$
3. Multiplication & Division (Left to Right):
   • First perform multiplication $4 \times 5 = 20$. $$\text{Expression: } 36 - 20 + 8 \div 4$$
   • Next perform division $8 \div 4 = 2$. $$\text{Expression: } 36 - 20 + 2$$
4. Addition & Subtraction (Left to Right):
   • First perform subtraction $36 - 20 = 16$. $$\text{Expression: } 16 + 2$$
   • Next perform addition $16 + 2 = \mathbf{18}$.

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2. Prime Numbers vs. Composite Numbers

Every positive integer greater than $1$ is classified as either prime or composite based on its number of factors.

Classification Rules:

• Prime Number: A whole number strictly greater than $1$ that has exactly two distinct factors: $1$ and itself.
  • The complete list of prime numbers less than $50$ is: $\mathbf{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}$.
  • Critical Fact: $2$ is the only even prime number. All other even numbers are divisible by $2$.
• Composite Number: A whole number strictly greater than $1$ that has more than two factors (it can be divided evenly by numbers other than $1$ and itself). Examples: $4, 6, 8, 9, 10, 12, 14, 15$.
• Special Cases ($0$ and $1$): $0$ and $1$ are neither prime nor composite. $1$ has only one factor ($1$), and $0$ has infinitely many factors.

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3. Prime Factorisation

The Fundamental Theorem of Arithmetic states that every composite number can be uniquely expressed as a product of prime numbers.

Factor Tree / Division Ladder Method for $180$:

1. Divide $180$ by the smallest prime factor ($2$): $180 \div 2 = 90$.
2. Divide $90$ by $2$: $90 \div 2 = 45$.
3. Divide $45$ by $3$: $45 \div 3 = 15$.
4. Divide $15$ by $3$: $15 \div 3 = 5$.
5. Divide $5$ by $5$: $5 \div 5 = 1$.

$$\text{Prime Factorisation of } 180 = 2 \times 2 \times 3 \times 3 \times 5 = \mathbf{2^2 \times 3^2 \times 5}$$

Factors, Multiples, HCF & LCM Applications

1. Factors and Multiples Definitions

• Factors: Whole numbers that divide evenly into a given number without leaving a remainder. For example, factors of $24$ are $\{1, 2, 3, 4, 6, 8, 12, 24\}$.
• Multiples: Numbers formed by multiplying a given number by any whole number. For example, multiples of $6$ are $\{6, 12, 18, 24, 30, 36, \dots\}$.

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2. Highest Common Factor (HCF) / Greatest Common Divisor (GCD)

The Highest Common Factor (HCF) of two or more numbers is the largest positive integer that divides each of the numbers without a remainder.

Method for Finding HCF Using Prime Factorisation:

1. Express each number as a product of prime powers.
2. Identify common prime factors shared by all numbers.
3. Take the lowest exponent for each common prime factor and multiply them.

Worked Example: Find HCF of $72$ and $108$

• Prime factorisation of $72$: $72 = 2^3 \times 3^2$.
• Prime factorisation of $108$: $108 = 2^2 \times 3^3$.
• Common prime bases: $2$ and $3$.
• Lowest power of base $2$: $2^2 = 4$.
• Lowest power of base $3$: $3^2 = 9$.
• $\text{HCF}(72, 108) = 2^2 \times 3^2 = 4 \times 9 = \mathbf{36}$.

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3. Least Common Multiple (LCM)

The Least Common Multiple (LCM) of two or more numbers is the smallest positive integer that is divisible by each of the numbers.

Method for Finding LCM Using Prime Factorisation:

1. Express each number as a product of prime powers.
2. Identify all prime bases present across any of the numbers.
3. Take the highest exponent for each prime factor and multiply them.

Worked Example: Find LCM of $72$ and $108$

• Prime factorisation of $72$: $2^3 \times 3^2$.
• Prime factorisation of $108$: $2^2 \times 3^3$.
• Highest power of base $2$: $2^3 = 8$.
• Highest power of base $3$: $3^3 = 27$.
• $\text{LCM}(72, 108) = 2^3 \times 3^3 = 8 \times 27 = \mathbf{216}$.

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4. Real-World Word Problem Applications of HCF and LCM

Understanding when to apply HCF versus LCM is critical for CPEA problem solving:

Decision Rule Table

Practical Problem Walkthroughs

Problem 1 (HCF Application): A school teacher has $48$ blue pencils and $72$ red pencils. She wishes to distribute them into identical gift bags containing equal numbers of blue and red pencils with zero pencils remaining. What is the maximum number of gift bags she can make?

• Analysis: Dividing items into maximum equal groups requires finding the HCF.
• Calculation: $\text{HCF}(48, 72)$.
  • $48 = 2^4 \times 3$
  • $72 = 2^3 \times 3^2$
  • $\text{HCF} = 2^3 \times 3 = 8 \times 3 = \mathbf{24 \text{ bags}}$.
• Details: Each bag contains $48 \div 24 = 2$ blue pencils and $72 \div 24 = 3$ red pencils.

Problem 2 (LCM Application): Three alarm clocks ring at intervals of $12$ minutes, $15$ minutes, and $20$ minutes respectively. If all three chime together at $8:00 \text{ AM}$, at what time will they next chime simultaneously?

• Analysis: Coinciding periodic events requires finding the LCM.
• Calculation: $\text{LCM}(12, 15, 20)$.
  • $12 = 2^2 \times 3$
  • $15 = 3 \times 5$
  • $20 = 2^2 \times 5$
  • $\text{LCM} = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60 \text{ minutes}$.
• Conclusion: $60 \text{ minutes} = 1 \text{ hour}$. Adding $1 \text{ hour}$ to $8:00 \text{ AM}$ gives $\mathbf{9:00 \text{ AM}}$.