📖 generic · CBSE Class 10 ENGLISH MEDIUM · MATHEMATICS · Page 1question

1.1 Introduction · Part 5

Chapter 1: REAL NUMBERS · MATHEMATICS

by the prime factorisation method. Solution : We have : = × and = × × = × . You can find HCF( , ) = and LCM( , ) = × × × = , as done in your earlier classes. Note that HCF( , ) = = Product of the smallest power of each common prime factor in the numbers.

LCM ( , ) = × × = Product of the greatest power of each prime factor , involved in the numbers . From the example above, you might have noticed that HCF( , ) × LCM( , ) = × . In fact, we can verify that for any two positive integers a and b , HCF ( a , b ) × LCM ( a , b ) = a × b . We can use this result to find the LCM of two positive integers, if we have already found the HCF of the two positive integers.

Example : Find the HCF of and by the prime factorisation method. Hence, find their LCM. Solution : The prime factorisation of and gives : = × , = × Therefore, the HCF of these two integers is = . Also, LCM ( , ) = HCF( , )   Example : Find the HCF and LCM of , and , using the prime factorisation method.

Solution : We have : = × , = × , = × × Here, and are the smallest powers of the common factors and , respectively. HCF ( , , ) = × = × = , and are the greatest powers of the prime factors , and

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