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\text{LCM} \times \text{HCF} = \text{product of the two numbers}.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

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Solution
Understand the Question
  • To find the HCF (Highest Common Factor), multiply the smallest power of each common prime factor.
  • To find the LCM (Least Common Multiple), multiply the greatest power of each prime factor involved in the numbers.
  • To verify the relationship for any two positive integers aa and bb: LCM(a,b)×HCF(a,b)=a×b\text{LCM}(a, b) \times \text{HCF}(a, b) = a \times b

(i) Find the LCM and HCF of 26 and 91 and verify that LCM×HCF=product of the two numbers\text{LCM} \times \text{HCF} = \text{product of the two numbers}.

Step 1 · Find HCF and LCM of 26 and 91

Diagram 1

Prime factorising the numbers 26=2×1326 = 2 \times 13 91=7×1391 = 7 \times 13

HCF(26,91)=13\text{HCF}(26, 91) = 13 LCM(26,91)=2×7×13=182\text{LCM}(26, 91) = 2 \times 7 \times 13 = 182

Step 2 · Verify LCM × HCF = Product of Numbers

Product of the two numbers Product of numbers=26×91=2366\text{Product of numbers} = 26 \times 91 = 2366

Product of HCF and LCM HCF×LCM=13×182=2366\text{HCF} \times \text{LCM} = 13 \times 182 = 2366

Since LCM×HCF=Product of numbers=2366\text{LCM} \times \text{HCF} = \text{Product of numbers} = 2366, the property is verified.

Answer

(i) HCF=13,LCM=182\text{HCF} = 13, \quad \text{LCM} = 182

(ii) Find the LCM and HCF of 510 and 92 and verify that LCM×HCF=product of the two numbers\text{LCM} \times \text{HCF} = \text{product of the two numbers}.

Step 1 · Find HCF and LCM of 510 and 92

Prime factorising the numbers 510=2×3×5×17510 = 2 \times 3 \times 5 \times 17

92=2×2×23=22×23\begin{aligned} 92 &= 2 \times 2 \times 23 \\ &= 2^2 \times 23 \end{aligned}

HCF(510,92)=21=2\text{HCF}(510, 92) = 2^1 = 2

LCM(510,92)=22×3×5×17×23=4×3×5×17×23=12×5×17×23=60×17×23=1020×23=23460\begin{aligned} \text{LCM}(510, 92) &= 2^2 \times 3 \times 5 \times 17 \times 23 \\ &= 4 \times 3 \times 5 \times 17 \times 23 \\ &= 12 \times 5 \times 17 \times 23 \\ &= 60 \times 17 \times 23 \\ &= 1020 \times 23 \\ &= 23460 \end{aligned}

Step 2 · Verify LCM × HCF = Product of Numbers

Product of the two numbers Product of numbers=510×92=46920\text{Product of numbers} = 510 \times 92 = 46920

Product of HCF and LCM HCF×LCM=2×23460=46920\text{HCF} \times \text{LCM} = 2 \times 23460 = 46920

Since LCM×HCF=Product of numbers=46920\text{LCM} \times \text{HCF} = \text{Product of numbers} = 46920, the property is verified.

Answer

(ii) HCF=2,LCM=23460\text{HCF} = 2, \quad \text{LCM} = 23460

(iii) Find the LCM and HCF of 336 and 54 and verify that LCM×HCF=product of the two numbers\text{LCM} \times \text{HCF} = \text{product of the two numbers}.

Step 1 · Find HCF and LCM of 336 and 54

Prime factorising the numbers

336=2×2×2×2×3×7=24×3×7\begin{aligned} 336 &= 2 \times 2 \times 2 \times 2 \times 3 \times 7 \\ &= 2^4 \times 3 \times 7 \end{aligned} 54=2×3×3×3=2×33\begin{aligned} 54 &= 2 \times 3 \times 3 \times 3 \\ &= 2 \times 3^3 \end{aligned} HCF(336,54)=21×31=2×3=6\begin{aligned} \text{HCF}(336, 54) &= 2^1 \times 3^1 \\ &= 2 \times 3 \\ &= 6 \end{aligned} LCM(336,54)=24×33×7=16×27×7=432×7=3024\begin{aligned} \text{LCM}(336, 54) &= 2^4 \times 3^3 \times 7 \\ &= 16 \times 27 \times 7 \\ &= 432 \times 7 \\ &= 3024 \end{aligned}

Step 2 · Verify LCM × HCF = Product of Numbers

Product of the two numbers Product of numbers=336×54=18144\text{Product of numbers} = 336 \times 54 = 18144

Product of HCF and LCM HCF×LCM=6×3024=18144\text{HCF} \times \text{LCM} = 6 \times 3024 = 18144

Since LCM×HCF=Product of numbers=18144\text{LCM} \times \text{HCF} = \text{Product of numbers} = 18144, the property is verified.

Answer

(iii) HCF=6,LCM=3024\text{HCF} = 6, \quad \text{LCM} = 3024

Common Mistakes
  • Power Confusion: Using the highest powers for HCF or the lowest powers for LCM. Remember: HCF uses the lowest powers of common factors, whereas LCM uses the highest powers of all factors involved.
  • Applicability of the Formula: The identity HCF×LCM=Product of numbers\text{HCF} \times \text{LCM} = \text{Product of numbers} is valid only for two numbers, not for three or more numbers.

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\text{LCM} \times \text{HCF} = \text{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\text{HCF}(306, 657) = 9, find LCM(306,657)\text{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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