NCERT Solutions
Class 10 - Mathematics - Chapter 1: REAL NUMBERS - EXERCISE 1.1
Question 3

Question. 3

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.

Answer:

(i) LCM = 420; HCF = 3

(ii) LCM = 11339; HCF = 1

(iii) LCM = 1800; HCF = 1

Detailed Answer with Explanation:

We will use the prime factorisation method. First, we write each number as a product of prime numbers.

HCF (Highest Common Factor) is the product of all prime factors that are common to all the given numbers, taken with their smallest powers.

LCM (Least Common Multiple) is the product of all prime factors that appear in any of the numbers, taken with their highest powers.

We now apply this to each group of numbers.

(i) 12, 15 and 21

First, write prime factorisation of each number.

For 12:

\(12 = 2^2 \times 3\)

For 15:

\(15 = 3 \times 5\)

For 21:

\(21 = 3 \times 7\)

Finding HCF

Look for prime factors that are common to all three numbers.

The prime factors are:

12 has \(2, 3\)

15 has \(3, 5\)

21 has \(3, 7\)

The only prime factor common to all three is \(3\).

So,

\(\text{HCF} = 3\)

Finding LCM

Now take all distinct prime factors appearing in any of the numbers.

These are \(2, 3, 5, 7\).

The highest powers among the three numbers are:

\(2^2\) from 12

\(3^1\) from each (all have at most one 3)

\(5^1\) from 15

\(7^1\) from 21

So,

\(\text{LCM} = 2^2 \times 3 \times 5 \times 7\)

Now multiply step by step.

\(2^2 = 4\)

\(4 \times 3 = 12\)

\(12 \times 5 = 60\)

\(60 \times 7 = 420\)

So,

\(\text{LCM} = 420\)

Final answers for (i)

\(\text{HCF} = 3\)

\(\text{LCM} = 420\)

(ii) 17, 23 and 29

All three numbers 17, 23 and 29 are prime numbers.

So their prime factorisations are:

\(17 = 17\)

\(23 = 23\)

\(29 = 29\)

Finding HCF

For the HCF, we look for prime factors that are common to all three numbers.

Since each number is a different prime, they have no common prime factor greater than 1.

So,

\(\text{HCF} = 1\)

Finding LCM

For the LCM of distinct primes, we simply multiply them together.

So,

\(\text{LCM} = 17 \times 23 \times 29\)

First multiply 17 and 23.

\(17 \times 23 = 391\)

Now multiply this result by 29.

\(391 \times 29 = 11339\)

So,

\(\text{LCM} = 11339\)

Final answers for (ii)

\(\text{HCF} = 1\)

\(\text{LCM} = 11339\)

(iii) 8, 9 and 25

Now factor each number into primes.

For 8:

Divide by 2 again and again.

\(8 = 2^3\)

For 9:

\(9 = 3^2\)

For 25:

\(25 = 5^2\)

Finding HCF

We look for prime factors that are common to all three numbers.

8 has only prime factor \(2\).

9 has only prime factor \(3\).

25 has only prime factor \(5\).

There is no prime factor that is common to all three.

So,

\(\text{HCF} = 1\)

Finding LCM

Now take all prime factors that appear in any of the numbers.

These are \(2, 3, 5\).

The highest powers are:

\(2^3\) from 8

\(3^2\) from 9

\(5^2\) from 25

So,

\(\text{LCM} = 2^3 \times 3^2 \times 5^2\)

Calculate step by step.

\(2^3 = 8\)

\(3^2 = 9\)

\(5^2 = 25\)

Now multiply these results.

\(8 \times 9 = 72\)

\(72 \times 25 = 1800\)

So,

\(\text{LCM} = 1800\)

Final answers for (iii)

\(\text{HCF} = 1\)

\(\text{LCM} = 1800\)

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