Real Numbers | Exercise 1.1

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

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Solution

We will find the prime factors of each number. Then we use these factors to find the HCF and LCM.

Step 1 — For 12, 15 and 21

Let's find the prime factors of each number.

12=2×2×312 = 2 \times 2 \times 3

12=22×3112 = 2^2 \times 3^1

15=3×515 = 3 \times 5

15=31×5115 = 3^1 \times 5^1

21=3×721 = 3 \times 7

21=31×7121 = 3^1 \times 7^1

To find the HCF, we take the lowest power of common prime factors. The only common prime factor is 3.

HCF(12,15,21)=3\boxed{\text{HCF}(12, 15, 21) = 3}

To find the LCM, we take the highest power of all prime factors. The prime factors are 2, 3, 5, 7.

LCM=22×31×51×71\text{LCM} = 2^2 \times 3^1 \times 5^1 \times 7^1

=4×3×5×7= 4 \times 3 \times 5 \times 7

=12×35= 12 \times 35

LCM(12,15,21)=420\boxed{\text{LCM}(12, 15, 21) = 420}

Diagram 1

Step 2 — For 17, 23 and 29

Let's find the prime factors of each number. These numbers are all prime numbers.

17=17117 = 17^1

23=23123 = 23^1

29=29129 = 29^1

To find the HCF, we take the lowest power of common prime factors. There are no common prime factors other than 1.

HCF(17,23,29)=1\boxed{\text{HCF}(17, 23, 29) = 1}

To find the LCM, we take the highest power of all prime factors. The prime factors are 17, 23, 29.

LCM=17×23×29\text{LCM} = 17 \times 23 \times 29

=391×29= 391 \times 29

LCM(17,23,29)=11339\boxed{\text{LCM}(17, 23, 29) = 11339}

Step 3 — For 8, 9 and 25

Let's find the prime factors of each number.

8=2×2×28 = 2 \times 2 \times 2

8=238 = 2^3

9=3×39 = 3 \times 3

9=329 = 3^2

25=5×525 = 5 \times 5

25=5225 = 5^2

To find the HCF, we take the lowest power of common prime factors. There are no common prime factors other than 1.

HCF(8,9,25)=1\boxed{\text{HCF}(8, 9, 25) = 1}

To find the LCM, we take the highest power of all prime factors. The prime factors are 2, 3, 5.

LCM=23×32×52\text{LCM} = 2^3 \times 3^2 \times 5^2

=8×9×25= 8 \times 9 \times 25

=72×25= 72 \times 25

LCM(8,9,25)=1800\boxed{\text{LCM}(8, 9, 25) = 1800}

Answer

(i) HCF = 3, LCM = 420 (ii) HCF = 1, LCM = 11339 (iii) HCF = 1, LCM = 1800

More questions in Exercise 1.1

Q1

Express each number as a product of its prime factors:

(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Q2

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54

Q3

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

Q4

Given that HCF (306, 657) = 9, find LCM (306, 657).

Q5

Check whether 6n6^n can end with the digit 0 for any natural number nn.

Q6

Explain why 7×11×13+137 \times 11 \times 13 + 13 and 7×6×5×4×3×2×1+57 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 are composite numbers.

Q7

There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

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