Page 62 - ICSE Math 6

Example 20: Find the HCF of 30, 24 and 48 using the common factor method.
Solution: Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.

Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
Common factors are 1, 2, 3, 6 and the largest common factor is 6.

∴ HCF of 30, 24 and 48 is 6.
Prime factorization method

Step 1: Find all the prime factors of all the given numbers.
Step 2: From the prime factors obtained, select the common prime factors.
Step 3: Multiply these common prime factors to obtain the HCF of the given numbers.

Example 21: Find the HCF of 50, 80 and 150 using the prime factorization method.
Solution: Since, 50 = 2 × 5 × 5 2 50 2 80 2 150
80 = 2 × 2 × 2 × 2 × 5 5 25 2 40 3 75
150 = 2 × 3 × 5 × 5 5 5 2 20 5 25
Common prime factors are 2 and 5. 1 2 10 5 5
∴ HCF = 2 × 5 = 10
5 5 1

Example 22: What is the HCF of two consecutive: 1
(a) numbers (b) odd numbers (c) even numbers (d) prime numbers
Solution: (a) The HCF of two consecutive numbers is 1, as 1 is the only common factor of two
consecutive numbers.
(b) The HCF of two consecutive odd numbers is 1, as 1 is the only common factor of
two consecutive odd numbers.
(c) The HCF of two consecutive even numbers is 2, as 1 and 2 are the only common
factors of two consecutive even numbers.
(d) The HCF of any two consecutive prime numbers is 1.

Example 23: The HCF of co-prime numbers 9 and 14 was calculated by the prime factorization
method as:
Prime factorization of 9 = 3 × 3 and prime factorization of 14 = 2 × 7.
Since, there is no common prime factor of 9 and 14, so the HCF of 9 and 14 is 0.
Is the answer correct? If not, what is the correct HCF?

Solution: No; the answer is not correct.
Apparently, there is no common prime factor in 9 and 14 but 1 is always a factor of
any number. Therefore, the HCF of 9 and 14 is 1 and not 0.

Division method
Step 1: Divide the bigger number by the smaller number.
Step 2: If there is no remainder left, then the smaller number is the HCF. If there is a remainder, then
take this remainder as the divisor and the previous divisor as the dividend. Divide the two.

Step 3: Perform division in this manner till there is no remainder left. The last divisor is the HCF
of the two given numbers.

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