Mathematics: Class 7 (Part 2) — Chapter 3: Finding Common Ground

Consider a room of size $12\text{ ft} \times 16\text{ ft}$ to be tiled with square tiles of the same size. To use the minimum number of tiles without cutting, the side length of the tile must be a whole number and a factor of both dimensions:

• Factors of $12$: $1, 2, 3, 4, 6, 12$
• Factors of $16$: $1, 2, 4, 8, 16$
• Common Factors: $1, 2, 4$

The largest possible tile size is the Highest Common Factor (HCF), which is $4\text{ ft}$.

The Highest Common Factor (HCF) or Greatest Common Divisor (GCD) of two or more numbers is the highest (or greatest) of their common factors.

Listing all factors can be tedious for large numbers. Prime Factorisation simplifies this process. Any number can be uniquely written as a product of prime numbers (except for the order of factors).

Example: Prime factorisation of $90$:

$$90 = 2 \times 3 \times 3 \times 5 = 2 \times 3^2 \times 5$$

Division Method for Prime Factorisation

We divide the number by prime factors successively until we reach $1$:

• For $105$: $105 \div 3 = 35 \implies 35 \div 5 = 7 \implies 7 \div 7 = 1$. Thus, $105 = 3 \times 5 \times 7$.
• For $30$: $30 \div 2 = 15 \implies 15 \div 3 = 5 \implies 5 \div 5 = 1$. Thus, $30 = 2 \times 3 \times 5$.

To find the HCF of two or more numbers using prime factorisation:

1. Find the prime factorisation of each number.
2. Identify the common prime factors.
3. For each common prime factor, take the minimum number of times it appears in any of the factorisations.
4. Multiply these common prime factors together.

Example 1: Find the HCF of $45$ and $75$.

• $45 = 3 \times 3 \times 5 = 3^2 \times 5$
• $75 = 3 \times 5 \times 5 = 3 \times 5^2$
• Common prime factors: $3$ (minimum exponent $1$) and $5$ (minimum exponent $1$).
• $\text{HCF} = 3 \times 5 = 15$.

Example 2: Find the HCF of $112$ and $84$.

• $112 = 2 \times 2 \times 2 \times 2 \times 7 = 2^4 \times 7$
• $84 = 2 \times 2 \times 3 \times 7 = 2^2 \times 3 \times 7$
• Common prime factors: $2$ (minimum exponent $2$) and $7$ (minimum exponent $1$).
• $\text{HCF} = 2^2 \times 7 = 2 \times 2 \times 7 = 28$.

Anshu uses strips of length $6\text{ cm}$ and Guna uses strips of length $8\text{ cm}$ to make torans of equal length. The possible lengths are common multiples of $6$ and $8$:

• Multiples of $6$: $6, 12, 18, 24, 30, 36, 42, 48, 54, \dots$
• Multiples of $8$: $8, 16, 24, 32, 40, 48, 56, 64, 72, \dots$
• Common Multiples: $24, 48, 72, \dots$

The smallest possible length is the Lowest Common Multiple (LCM), which is $24\text{ cm}$.

The Lowest Common Multiple (LCM) of two or more given numbers is the lowest (or smallest or least) of their common multiples.

To find the LCM of two or more numbers using prime factorisation:

1. Find the prime factorisation of each number.
2. Identify all the prime factors that appear in any of the numbers.
3. For each prime factor, take the maximum number of times it appears in any of the factorisations.
4. Multiply these prime factors together.

Example: Find the LCM of $96$ and $360$.

• $96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3$
• $360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 2^3 \times 3^2 \times 5$
• Prime factors involved: $2, 3, 5$.
• Maximum occurrences: $2$ appears $5$ times, $3$ appears $2$ times, $5$ appears $1$ time.
• $\text{LCM} = 2^5 \times 3^2 \times 5^1 = 32 \times 9 \times 5 = 1440$.

• Factor-Multiple Relationship: If one number is a factor of another, then the smaller number is their HCF and the larger number is their LCM.
  • For $n$ and $5n$, $\text{HCF} = n$ and $\text{LCM} = 5n$.
• Scaling Property: If both numbers are multiplied by a constant $k$, their HCF and LCM are also multiplied by $k$.
  • $\text{HCF}(k \cdot a, k \cdot b) = k \cdot \text{HCF}(a, b)$
  • $\text{LCM}(k \cdot a, k \cdot b) = k \cdot \text{LCM}(a, b)$

For any two positive integers $a$ and $b$, the product of their HCF and LCM is equal to the product of the two numbers:

$$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$$

Example: Verify for $105$ and $95$.

• $105 = 3 \times 5 \times 7$, $95 = 5 \times 19$
• $\text{HCF} = 5$
• $\text{LCM} = 3 \times 5 \times 7 \times 19 = 1995$
• $\text{HCF} \times \text{LCM} = 5 \times 1995 = 9975$
• Product of numbers $= 105 \times 95 = 9975$

We can find both HCF and LCM simultaneously using a common division table. We divide both numbers by common factors (not necessarily prime) until no common factors remain.

Example for $300$ and $150$:

• Divide by $50$: quotients are $6$ and $3$.
• Divide by $3$: quotients are $2$ and $1$.
• No common factors remain between $2$ and $1$.
• $\text{HCF} = 50 \times 3 = 150$
• $\text{LCM} = 50 \times 3 \times 2 \times 1 = 300$