NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 3(iii)

NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 3(iii)

Question:

Find the LCM and HCF of the following integers by applying the prime factorisation method: 8, 9, and 25

Given:

Numbers: 8, 9, 25

To Find:

LCM and HCF of 8, 9, and 25

Formula:

  • Prime Factorisation: Express each number as a product of its prime factors.
  • HCF: Product of the lowest powers of all common prime factors.
  • LCM: Product of the highest powers of all prime factors present in any number.

Solution:

Step 1: Write each number as a product of its prime factors.

$$ 8 = 2 \times 2 \times 2 = 2^3 $$ $$ 9 = 3 \times 3 = 3^2 $$ $$ 25 = 5 \times 5 = 5^2 $$

Step 2: Find the HCF (Highest Common Factor).

Check for common prime factors in all three numbers.

There is no prime factor common to 8, 9, and 25.

$$ \gcd(8,\,9,\,25) = 1 $$

Step 3: Find the LCM (Lowest Common Multiple).

List all prime factors present in any of the numbers: 2, 3, 5.

Take the highest power of each prime:

  • 2: \(2^3\) (from 8)
  • 3: \(3^2\) (from 9)
  • 5: \(5^2\) (from 25)
$$ \mathrm{lcm}(8,\,9,\,25) = 2^3 \times 3^2 \times 5^2 $$

Calculate each power:

$$ 2^3 = 8 $$ $$ 3^2 = 9 $$ $$ 5^2 = 25 $$

Multiply the results:

$$ 8 \times 9 = 72 $$ $$ 72 \times 25 = 1800 $$

Therefore, the LCM of 8, 9, and 25 is 1800.

Result:

HCF: $$ \gcd(8,\,9,\,25) = 1 $$
LCM: $$ \mathrm{lcm}(8,\,9,\,25) = 1800 $$

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