Page 57 - Maths Class 05

HCF by Factorisation Method

Example : Find the HCF of 120 and 165 by factorisation method.

Solution : First of all, factorise the given numbers. Then, take all the common factors
and multiply them. The product will be the HCF of the given numbers.

To factorise the numbers, use the divisibility properties of numbers. Now,
observe the solution given below :

2 120 3 165
2 60 5 55

2 30 11 11

3 15 1
5 5

1

120 = 2 × 2 × 2 × 3 × 5 165 = 3 × 5 × 11

Common factors are 3 and 5.

Now, HCF = 3 × 5 = 15

Example : Find the HCF of 432, 648 and 1620.
Solution :
2 432 2 648 2 1620
2 216 2 324 2 810 432 = 2 × 2 × 2 × 2 × 3 × 3 × 3

2 108 2 162 3 405 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3

2 54 3 81 3 135 1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5
3 27 3 27 3 45 HCF = 2 × 2 × 3 × 3 × 3 = 108

3 9 3 9 3 15

3 3 3 3 5 5
1 1 1

HCF By Division Method

Example : Find HCF of 144 and 204.
Solution : First of all, divide the large number by smaller one, then the remainder is

treated as divisor and the divisor as dividend. Divide the first divisor by
first remainder. Similarly, divide second divisor by second remainder.

Continue the process till remainder becomes zero. Last divisor is HCF of
given numbers.

Mathematics-5 57