Question 9

Step 1 · Analyze Examples and Find LCM

Example 1: For $6$ and $9$:

• Multiples of $6$: $6, 12, \mathbf{18}, 24, \dots$
• Multiples of $9$: $9, \mathbf{18}, 27, \dots$ $\text{LCM}(6, 9) = 18$

Example 2: For $9$ and $12$:

• Multiples of $9$: $9, 18, 27, \mathbf{36}, \dots$
• Multiples of $12$: $12, 24, \mathbf{36}, \dots$ $\text{LCM}(9, 12) = 36$

$\boxed{\text{LCM}(6, 9) = 18, \quad \text{LCM}(9, 12) = 36}$

Step 2 · General Explanation

Since both numbers are divisible by $3$, $3$ is a factor of each number. The LCM must contain all prime factors of both numbers, so it must also contain $3$ as a factor.

$\text{General Statement: The LCM of two multiples of 3 is always a multiple of 3.}$

Step 1 · Analyze Examples and Find LCM

Example 1: For $2$ and $4$:

• Multiples of $2$: $2, \mathbf{4}, 6, \dots$
• Multiples of $4$: $\mathbf{4}, 8, \dots$ $\text{LCM}(2, 4) = 4 = \dfrac{2 \times 4}{2}$

Example 2: For $6$ and $8$:

• Multiples of $6$: $6, 12, 18, \mathbf{24}, \dots$
• Multiples of $8$: $8, 16, \mathbf{24}, \dots$ $\text{LCM}(6, 8) = 24 = \dfrac{6 \times 8}{2}$

Example 3: For $10$ and $12$:

• Multiples of $10$: $10, 20, 30, 40, 50, \mathbf{60}, \dots$
• Multiples of $12$: $12, 24, 36, 48, \mathbf{60}, \dots$ $\text{LCM}(10, 12) = 60 = \dfrac{10 \times 12}{2}$

$\boxed{\text{LCM}(2, 4) = 4, \quad \text{LCM}(6, 8) = 24, \quad \text{LCM}(10, 12) = 60}$

Step 2 · General Derivation

Let the two consecutive even numbers be $2n$ and $2n+2 = 2(n+1)$.

Since $n$ and $n+1$ are consecutive integers, they are co-prime with $\text{HCF}(n, n+1) = 1$. Thus, $\text{HCF}(2n, 2n+2) = 2$.

Using the formula $\text{LCM}(a, b) = \dfrac{a \times b}{\text{HCF}(a, b)}$:

$\text{General Statement: The LCM of two consecutive even numbers } 2n \text{ and } 2n+2 \text{ is half of their product, } 2n^2 + 2n.$

Step 1 · Analyze Examples and Find LCM

Example 1: For $7$ and $8$:

• Multiples of $7$: $7, 14, 21, 28, 35, 42, 49, \mathbf{56}, \dots$
• Multiples of $8$: $8, 16, 24, 32, 40, 48, \mathbf{56}, \dots$ $\text{LCM}(7, 8) = 56 = 7 \times 8$

Example 2: For $9$ and $10$:

• Multiples of $9$: $9, 18, 27, 36, 45, 54, 63, 72, 81, \mathbf{90}, \dots$
• Multiples of $10$: $10, 20, 30, 40, 50, 60, 70, 80, \mathbf{90}, \dots$ $\text{LCM}(9, 10) = 90 = 9 \times 10$

$\boxed{\text{LCM}(7, 8) = 56, \quad \text{LCM}(9, 10) = 90}$

Step 2 · General Explanation

Any two consecutive integers $a$ and $b$ share no common factor other than $1$, so $\text{HCF}(a, b) = 1$.

$\text{General Statement: The LCM of two consecutive numbers is equal to their product.}$

Step 1 · Analyze Examples and Find LCM

Example 1: For $4$ and $9$:

• Factors of $4$: $1, 2, 4$
• Factors of $9$: $1, 3, 9$
• $\text{HCF}(4, 9) = 1$
• Multiples of $4$: $4, 8, 12, 16, 20, 24, 28, 32, \mathbf{36}, \dots$
• Multiples of $9$: $9, 18, 27, \mathbf{36}, \dots$ $\text{LCM}(4, 9) = 36 = 4 \times 9$

Example 2: For $5$ and $8$:

• Factors of $5$: $1, 5$
• Factors of $8$: $1, 2, 4, 8$
• $\text{HCF}(5, 8) = 1$
• Multiples of $5$: $5, 10, 15, 20, 25, 30, 35, \mathbf{40}, \dots$
• Multiples of $8$: $8, 16, 24, 32, \mathbf{40}, \dots$ $\text{LCM}(5, 8) = 40 = 5 \times 8$

$\boxed{\text{LCM}(4, 9) = 36, \quad \text{LCM}(5, 8) = 40}$

Step 2 · General Explanation

By definition, co-prime numbers have an $\text{HCF} = 1$.

$\text{General Statement: The LCM of two co-prime numbers is equal to their product.}$