Page 49 - Math Skill - 5
P. 49
Factors and Multiples 47
Example 2: Find the HCF of 160 and 180.
Solution: Factorisation of 160 Factorisation of 180
2 160 2 180
2 80
2 40 2 90
2 20 Prime factors of 160 3 45 Prime factors of 180
2 10 = 2 × 2 × 2 × 2 × 2 × 5 3 15 = 2 × 2 × 3 × 3 × 5
5 5 5 5
1 1
Common factors = 2, 2, 5
Product of the common factors = 2 × 2 × 5 = 20
HCF of 160 and 180 is 20.
Finding HCF by Long Division Method
HCF of two or more numbers can also be found by the process of repeated division.
The following shows the steps which one must follow for finding HCF by long division method:
Divide the larger number by the smaller one.
Divide the first divisor by the first remainder.
Divide the second divisor by the second remainder.
Continue the process till the remainder turns zero.
The last divisor is the required HCF of the given numbers.
Let’s Attempt
Example 1: Find the HCF of 48 and 18 by long division method.
Solution: 18 48 2
– 36
12 18 1
– 12
Last divisor 6 12 2
– 12
0 \ HCF of 18 and 48 is 6.
Example 2: Find the HCF of 136, 170 and 255.
Solution: First we find the HCF of 136 and 170. Now, we find the HCF of 34 and 255
136 170 1 34 255 7
– 136 – 238
34 136 4 17 34 4
– 136 – 34
0 0
HCF of 136 and 170 is 34. HCF of 34 and 255 is 17.
\ HCF of 136, 170, 255 is 17.
