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) 2^{5} × 5^{4} × 7^{2} × 13^{6} and 2^{3} × 5^{6} × 7 × 17^{3}

(ii) a^{5} × b^{2} × c^{2} × d^{5} and a^{7} × b^{3} × e × f^{3} where a, b, c, d, e and f are prime .

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

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

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 2^{3}, 5^{4}, and 7^{1}

∴ HCF of 2^{5} × 5^{4} × 7^{2} × 13^{6} and 2^{3} × 5^{6} × 7 × 17^{3} is **2 ^{3} × 5^{4} × 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 2^{5}, 5^{6}, 7^{2}, 13^{6}, and 17^{3} respectively.

∴ LCM of 2^{5} × 5^{4} × 7^{2} × 13^{6} and 2^{3} × 5^{6} × 7 × 17^{3} is **2 ^{5} × 5^{6} × 7^{2} × 13^{6} × 17^{3}**

** 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 a^{5} and b^{2} respectively.

∴ HCF of a^{5} × b^{2} × c^{2} × d^{5} and a^{7} × b^{3} × e × f^{3} is **a ^{5} × b^{2}**

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 a^{7}, b^{3}, c^{2}, d^{5}, e^{1}, and f^{3} respectively.

∴ LCM of a^{5} × b^{2} × c^{2} × d^{5} and a^{7} × b^{3} × e × f^{3} is **a ^{7} × b^{3} × c^{2} × d^{5} × e × f^{3}**

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