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Chapters For Class X- CBSE


H.C.F and L.C.M by prime factorisation method

Example : Find the LCM and HCF of 60 and 200 by the prime factorisation method.


60 2 30 2 15 3 5 60=2×2×3×5 200 2 100 2 50 2 25 5 5 200=2×2×2×5×5
HCF(60, 200) = 22$×5$ =$20$( Product of the smallest power of each common prime factor in the numbers).

LCM (60, 200) = 23×3×52=$600$ (Product of the greatest power of each prime factor, involved in the numbers.)

For any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b.
Example : Find the LCM and HCF of 36 and 81 by the prime factorisation method.

36 3 12 3 4 2 2 36=3×3×2×2 81 3 27 3 9 3 3 81=3×3×3×3
HCF(36,81) = 32 =$9$( Product of the smallest power of each common prime factor in the numbers).

LCM (36,81) = 34×22=324 (Product of the greatest power of each prime factor, involved in the numbers.)

Example : Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM.

 Solution : The prime factorisation of 96 and 404 gives :

96 = 25 × 3
404 = 22 × 101

Therefore, the HCF of these two integers (96,404) is 22 = 4.

Also, HCF (a, b) × LCM (a, b) = a × b.

4 × LCM(96,404)=96 × 404

Therefore , LCM(96,404) = (96 × 404 )/4 = 9696


Example : Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method.

Solution : We have :

6 = 2 × 3,
72 = 23 × 32,
120 = 23 × 3 × 5

Here, 21 and 31 are the smallest powers of the common factors 2 and 3, respectively.

So,HCF (6, 72, 120) = 21 × 31 = 2 × 3 = 6

23, 32 and 51 are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers.

So, LCM (6, 72, 120) = 23 × 32 × 51 = 360

Remark : Notice, 6 × 72 × 120 ≠ HCF (6, 72, 120) × LCM (6, 72, 120).
So, the product of three numbers is not equal to the product of their HCF and LCM.