# Ex.1.2 Q3 Real Numbers Solution - NCERT Maths Class 10

## Question

Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) \(12, 15\) and \(21\)

(ii) \(17, 23\) and \(29 \)

(iii) \( 8, 9\) and \(25\)

## Text Solution

**What is known?**

(i) \(12, 15\) and \(21\)

(ii) \(17, 23\) and \(29 \)

(iii) \( 8, 9\) and \(25\)

**What is unknown?**

The LCM and HCF of the given integers by applying the prime factorisation method.

**Reasoning:**

To solve this question, follow the following steps-

- First find the prime factors of the given integers.
- Find the HCF of the given pair of integers i.e. product of smallest power of each prime factor, involved in the number.
- Lastly, Find the LCM of the given pair of integers i.e. product of greatest power of each prime factor, involved in the number.

**Steps:**

(i) \(12, 15\) and \(21\)

Prime factors of \(12\)

\[\begin{align}&= {2 \times 2 \times 3}\\&={{2^2} \times 3}\end{align}\]

Prime factors of \(15= 3 \times 5\)

Prime factors of \(21=2 \times 2 \times 3\)

HCF of \(12,15\) and \(21= {3}\)

LCM of \(12,15\) and \(21\) \[\begin{align}&= {{2^2} \times 3 \times 5 \times 7}\\ &={ 420}\end{align}\]

(ii) \(17, 23\) and \(29 \)

Prime factors of \(17= 17\,\times\,1\)

Prime factors of \(23=23\,\times\,1\)

Prime factors of \(29 = 29 \times 1\)

HCF of \(7,\,23\) and \( 29={1}\)

LCM of \(17,\,23\) and \(29\)

\[\begin{align}&= 17 \times 23 \times 29\\&= 11339\end{align}\]

(iii) \( 8, 9\) and \(25\)

Prime factors of \(8\)

\[\begin{align}&= 2 \times 2 \times 2\\ &= {2^3}\end{align}\]

Prime factors of \(9\)

\[\begin{align}&= 3 \times 3\\ &= {3^2}\end{align}\]

Prime factors of \(25\)

\[\begin{align}&= 5 \times 5\\ &= {\text{ }}{5^2}\end{align}\]

HCF of \(8, 9\) and \(25 = 1\)

LCM of \(8,\,9\) and \(25\)

\[\begin{align} &=2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 \\ &=1800 \end{align}\]