# Extra Questions For Class 10 Maths Chapter 1 | Q6

#### Real Numbers | Prime Factorization | Find LCM and HCF

This CBSE class 10 Maths Extra Practice Question is from the chapter Real Numbers. This question is about computing HCF and LCM of numbers using the prime factorisation method.

Question 6 : Find LCM and HCF of the following:
(i) 25 × 54 × 72 × 136 and 23 × 56 × 7 × 173
(ii) a5 × b2 × c2 × d5 and a7 × b3 × e × f3 where a, b, c, d, e and f are prime .

## NCERT Solution to Class 10 Maths

#### How to compute HCF by Prime Factorisation Method?

HCF of two numbers is the product of LOWEST power of COMMON primes found in the numbers.

#### How to find LCM by Prime Factorisation Method?

LCM of two numbers is the product of HIGHEST power of ALL primes found in the numbers.

#### (i)LCM and HCF of 25 × 54 × 72 × 136 and 23 × 56 × 7 × 173

Notice that the two numbers are already prime factorized. That makes life really easy.
Let us compute HCF first.
The common primes in the two numbers are 2, 5, and 7.
The lowest power of 2, 5, and 7 are respectively 23, 54, and 71
∴ HCF of 25 × 54 × 72 × 136 and 23 × 56 × 7 × 173 is 23 × 54 × 7

Let us compute LCM next.
The primes found in the two numbers are 2, 5, 7, 13, and 17.
The highest power of 2, 5, 7, 13, and 17 are 25, 56, 72, 136, and 173 respectively.
∴ LCM of 25 × 54 × 72 × 136 and 23 × 56 × 7 × 173 is 25 × 56 × 72 × 136 × 173

#### (ii) Find the LCM and HCF of a5 × b2 × c2 × d5 and a7 × b3 × e × f3

Given: a, b, c, d, e, and f are prime numbers.
Let us compute HCF first.
The common primes in the two numbers are a and b.
The lowest power of a and b in the two numbers are a5 and b2 respectively.
∴ HCF of a5 × b2 × c2 × d5 and a7 × b3 × e × f3 is a5 × b2

Let us compute LCM next.
The primes found in the two numbers are a, b, c, d, e, and f.
The highest power of a, b, c, d, e, and f are a7, b3, c2, d5, e1, and f3 respectively.
∴ LCM of a5 × b2 × c2 × d5 and a7 × b3 × e × f3 is a7 × b3 × c2 × d5 × e × f3

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