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

NCERT Class X Chapter 1: Real Numbers

Question:

Find the LCM and HCF of 12, 15, and 21 by applying the prime factorization method.

Given:

Numbers: 12, 15, 21

To Find:

LCM and HCF of 12, 15, and 21

Formula:

  • HCF (Highest Common Factor): The product of the smallest powers of all common prime factors in the numbers.
  • LCM (Least Common Multiple): The product of the greatest powers of all prime numbers present in any of the numbers.

Solution:

Step 1: Write the prime factorization of each number.

$$ 12 = 2 \times 2 \times 3 = 2^2 \times 3 $$ $$ 15 = 3 \times 5 $$ $$ 21 = 3 \times 7 $$

Step 2: Express each number using exponents for the prime factors.

  • $$ 12 = 2^2 \times 3^1 $$
  • $$ 15 = 3^1 \times 5^1 $$
  • $$ 21 = 3^1 \times 7^1 $$

Step 3: Find the HCF by taking the lowest power of each common prime factor.

The only common prime factor is 3. The lowest power is 1.

$$ \text{HCF}(12, 15, 21) = 3^1 = 3 $$

Step 4: Find the LCM by taking the highest power of each prime present in any number.

  • 2: highest power is 2 (from 12)
  • 3: highest power is 1 (from all)
  • 5: highest power is 1 (from 15)
  • 7: highest power is 1 (from 21)
$$ \text{LCM}(12, 15, 21) = 2^2 \times 3^1 \times 5^1 \times 7^1 $$

Step 5: Calculate the value of the LCM step by step.

$$ 2^2 = 4 $$ $$ 4 \times 3 = 12 $$ $$ 12 \times 5 = 60 $$ $$ 60 \times 7 = 420 $$ $$ \text{LCM}(12, 15, 21) = 420 $$

Result:

  • HCF(12, 15, 21) = 3
  • LCM(12, 15, 21) = 420
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