Question 7

We will find the Highest Common Factor (HCF) for different pairs of numbers. We will look at examples first. Then we will make a general statement. Finally, we will explain why it is true.

Step 1 — Two consecutive even numbers

Let us consider some examples. The numbers are (2, 4). Factors of 2 are 1, 2. Factors of 4 are 1, 2, 4. The common factors are 1, 2. The HCF of (2, 4) is 2.

Let us consider another pair. The numbers are (6, 8). Factors of 6 are 1, 2, 3, 6. Factors of 8 are 1, 2, 4, 8. The common factors are 1, 2. The HCF of (6, 8) is 2.

Let us consider one more pair. The numbers are (10, 12). Factors of 10 are 1, 2, 5, 10. Factors of 12 are 1, 2, 3, 4, 6, 12. The common factors are 1, 2. The HCF of (10, 12) is 2.

General Statement: The HCF of any two consecutive even numbers is always 2.

Reason: Let the two consecutive even numbers be $2n$ and $2n+2$. Both numbers are even. So, 2 is a common factor for them. The difference between these numbers is: $(2n+2) - 2n$ $= 2$ Any common factor of two numbers must divide their difference. So, the HCF must divide 2. The factors of 2 are 1 and 2. Since 2 is a common factor, the HCF cannot be 1. So, the HCF must be 2.

$\boxed{\text{HCF of two consecutive even numbers is 2.}}$

Step 2 — Two consecutive odd numbers

Let us consider some examples. The numbers are (3, 5). Factors of 3 are 1, 3. Factors of 5 are 1, 5. The common factor is 1. The HCF of (3, 5) is 1.

Let us consider another pair. The numbers are (7, 9). Factors of 7 are 1, 7. Factors of 9 are 1, 3, 9. The common factor is 1. The HCF of (7, 9) is 1.

Let us consider one more pair. The numbers are (11, 13). Factors of 11 are 1, 11. Factors of 13 are 1, 13. The common factor is 1. The HCF of (11, 13) is 1.

General Statement: The HCF of any two consecutive odd numbers is always 1.

Reason: Let the two consecutive odd numbers be $2n-1$ and $2n+1$. Both numbers are odd. So, 2 is not a common factor for them. The difference between these numbers is: $(2n+1) - (2n-1)$ $= 2$ Any common factor of two numbers must divide their difference. So, the HCF must divide 2. The factors of 2 are 1 and 2. Since 2 is not a common factor, the HCF cannot be 2. So, the HCF must be 1.

$\boxed{\text{HCF of two consecutive odd numbers is 1.}}$

Step 3 — Two even numbers

Let us consider some examples. The numbers are (4, 10). Factors of 4 are 1, 2, 4. Factors of 10 are 1, 2, 5, 10. The common factors are 1, 2. The HCF of (4, 10) is 2.

Let us consider another pair. The numbers are (8, 12). Factors of 8 are 1, 2, 4, 8. Factors of 12 are 1, 2, 3, 4, 6, 12. The common factors are 1, 2, 4. The HCF of (8, 12) is 4.

Let us consider one more pair. The numbers are (14, 20). Factors of 14 are 1, 2, 7, 14. Factors of 20 are 1, 2, 4, 5, 10, 20. The common factors are 1, 2. The HCF of (14, 20) is 2.

General Statement: The HCF of any two even numbers is always an even number.

Reason: Let the two even numbers be $2m$ and $2n$. Both numbers are even. This means both numbers have 2 as a factor. So, 2 is always a common factor. The HCF is the greatest common factor. Since 2 is a common factor, the HCF must be a multiple of 2. Any multiple of 2 is an even number. So, the HCF must be an even number.

$\boxed{\text{HCF of two even numbers is always an even number.}}$

Step 4 — Two consecutive numbers

Let us consider some examples. The numbers are (7, 8). Factors of 7 are 1, 7. Factors of 8 are 1, 2, 4, 8. The common factor is 1. The HCF of (7, 8) is 1.

Let us consider another pair. The numbers are (14, 15). Factors of 14 are 1, 2, 7, 14. Factors of 15 are 1, 3, 5, 15. The common factor is 1. The HCF of (14, 15) is 1.

Let us consider one more pair. The numbers are (20, 21). Factors of 20 are 1, 2, 4, 5, 10, 20. Factors of 21 are 1, 3, 7, 21. The common factor is 1. The HCF of (20, 21) is 1.

General Statement: The HCF of any two consecutive numbers is always 1.

Reason: Let the two consecutive numbers be $n$ and $n+1$. The difference between these numbers is: $(n+1) - n$ $= 1$ Any common factor of two numbers must divide their difference. So, the HCF must divide 1. The only factor of 1 is 1. So, the HCF must be 1.

$\boxed{\text{HCF of two consecutive numbers is 1.}}$

Step 5 — Two co-prime numbers

Let us consider some examples. The numbers are (4, 9). Factors of 4 are 1, 2, 4. Factors of 9 are 1, 3, 9. The common factor is 1. The HCF of (4, 9) is 1.

Let us consider another pair. The numbers are (5, 8). Factors of 5 are 1, 5. Factors of 8 are 1, 2, 4, 8. The common factor is 1. The HCF of (5, 8) is 1.

General Statement: The HCF of two co-prime numbers is always 1.

Reason: Co-prime numbers are defined in a special way. They are numbers that have only 1 as a common factor. This means they share no other common factors. So, their Highest Common Factor must be 1.

$\boxed{\text{HCF of two co-prime numbers is 1.}}$

Answer

(a) The HCF of any two consecutive even numbers is 2. (b) The HCF of any two consecutive odd numbers is 1. (c) The HCF of two even numbers is always an even number (at least 2). (d) The HCF of any two consecutive numbers is 1. (e) The HCF of two co-prime numbers is always 1.