Real Numbers | Exercise 1.1

Question 2

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

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Solution

We will use prime factorization to find the LCM and HCF. Then we will verify the product rule.

Step 1 — Find LCM and HCF for 26 and 91

Let's find the prime factors.

26=2×1326 = 2 \times 13

91=7×1391 = 7 \times 13

We find common prime factors. We take the lowest power.

HCF=13\boxed{\text{HCF} = 13}

We find all prime factors. We take the highest power.

LCM=2×7×13\text{LCM} = 2 \times 7 \times 13

LCM=182\boxed{\text{LCM} = 182}

Let's multiply the given numbers.

Product of numbers=26×91\text{Product of numbers} = 26 \times 91

2366\boxed{2366}

Let's multiply the HCF and LCM.

HCF×LCM=13×182\text{HCF} \times \text{LCM} = 13 \times 182

2366\boxed{2366}

Both results are the same. The property is verified.

Diagram 1

Step 2 — Find LCM and HCF for 510 and 92

Let's find the prime factors.

510=2×3×5×17510 = 2 \times 3 \times 5 \times 17

92=2×2×23=22×2392 = 2 \times 2 \times 23 = 2^2 \times 23

We find common prime factors. We take the lowest power.

HCF=2\boxed{\text{HCF} = 2}

We find all prime factors. We take the highest power.

LCM=22×3×5×17×23\text{LCM} = 2^2 \times 3 \times 5 \times 17 \times 23

=4×3×5×17×23= 4 \times 3 \times 5 \times 17 \times 23

=12×5×17×23= 12 \times 5 \times 17 \times 23

=60×17×23= 60 \times 17 \times 23

=1020×23= 1020 \times 23

LCM=23460\boxed{\text{LCM} = 23460}

Let's multiply the given numbers.

Product of numbers=510×92\text{Product of numbers} = 510 \times 92

46920\boxed{46920}

Let's multiply the HCF and LCM.

HCF×LCM=2×23460\text{HCF} \times \text{LCM} = 2 \times 23460

46920\boxed{46920}

Both results are the same. The property is verified.

Step 3 — Find LCM and HCF for 336 and 54

Let's find the prime factors.

336=2×2×2×2×3×7=24×3×7336 = 2 \times 2 \times 2 \times 2 \times 3 \times 7 = 2^4 \times 3 \times 7

54=2×3×3×3=2×3354 = 2 \times 3 \times 3 \times 3 = 2 \times 3^3

We find common prime factors. We take the lowest power.

HCF=21×31\text{HCF} = 2^1 \times 3^1

=2×3= 2 \times 3

HCF=6\boxed{\text{HCF} = 6}

We find all prime factors. We take the highest power.

LCM=24×33×7\text{LCM} = 2^4 \times 3^3 \times 7

=16×27×7= 16 \times 27 \times 7

=432×7= 432 \times 7

LCM=3024\boxed{\text{LCM} = 3024}

Let's multiply the given numbers.

Product of numbers=336×54\text{Product of numbers} = 336 \times 54

18144\boxed{18144}

Let's multiply the HCF and LCM.

HCF×LCM=6×3024\text{HCF} \times \text{LCM} = 6 \times 3024

18144\boxed{18144}

Both results are the same. The property is verified.

Answer

(i) HCF = 13, LCM = 182. Product of numbers = 2366, HCF × LCM = 2366. Verified. (ii) HCF = 2, LCM = 23460. Product of numbers = 46920, HCF × LCM = 46920. Verified. (iii) HCF = 6, LCM = 3024. Product of numbers = 18144, HCF × LCM = 18144. Verified.

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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