### Chapter 1: Real Numbers

Q
##### Real Numbers Solutions

Question:

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

Solution:

(i) 26 and 91

The prime factors of 26 and 91 are :

26 = 2 x 13

91= 7 x 13

We know that LCM of two or more numbers = product of the greatest power of each prime factor involved in the numbers, with the highest power.

$\therefore$ L.C.M (26,91) = 2 x 7 x 13

Hence, L.C.M of (26, 91) = 182

H.C.F of (26,  91) = 13

Since, Highest Common Factor of two or more numbers = Product of the smallest Power of each common prime factor involved in the numbers.

We know that L.C.M x H.C.F = First number x Second number

182 x 13 = 26 x 91

2366 = 2366. Verified

(ii) 510 and 92

The prime factors of 510 and 92 are :

510 = 2 x 3 x 5 x 17

92 = 2 x 2 x 23 = 2² × 23

We know that LCM of two or more numbers = product of the greatest power of each prime factor involved in the numbers, with the highest power.

$\therefore$ L.C.M ( 510 ,92) = 2 x 2 x 3 x 5 x 23 x 17

L.C.M (510 ,92) = 23460

H.C.F (510 , 92) = 2

Since, HCF of two or more numbers= Product of the smallest Power of each common prime factor involved in the numbers.

We know that L.C.M x H.C.F = First Number x Second Number

23460 x 2 = 510 x 92

46920 = 46920. verified.

(iii) 336 and 54

The prime factors of 336 and 54 are :

336 = 2 x 2 x 2 x 2 x 3 x 7 = 2⁴ × 3¹ × 7¹

54 = 2 x 3 x 3 x 3 = 2¹ × 3³

We know that LCM of two or more numbers = product of the greatest power of each prime factor involved in the numbers, with the highest power.

$\therefore$ L.C.M (336 , 54) = 2⁴ × 3³× 7 = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 7

Hence, L.C.M( 336 , 54) = 3024

H.C.F (336 , 54) = 2 × 3

H.C.F (336 , 54) = 6

Since, HCF of two or more numbers = Product of the smallest Power of each common prime factor involved in the numbers.

We know that L.C.M x H.C.F = First Number x Second Number

3024 x 6 = 336 x 54

18144 = 18144. Verified

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