Chapter 1: Real Numbers

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Real Numbers Solutions

Question:

Find the LCM and HCF of the following integers by applying the prime factorization method. (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25

Answer:

(i) 12, 15 and 21

The prime factors of 15, 12 and 21 are :

12 = 2 x 2 x 3 = 2² × 3¹

15 = 3 x 5 = 3¹ × 5¹

21 = 3 x 7 = 3¹ × 7¹

L.C.M of 12, 15 and 21 =2² × 3¹ × 5¹ × 7¹ =  2 x 2 x 3 x 5 x 7 =  420

Hence, L.C.M of 12, 15 and 21 = 420

 

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

 

H.C.F of 12, 15 and 21 = 3

 

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

 

Hence, H.C.F( 12, 15 and 21) = 3 & L.C.M (12, 15 and 21)  420.

 

(ii) 17, 23 and 29              

The prime factors of 17, 23 and 29 are:  

17 = 1 x 17

23 = 1 x 23

29 = 1 x 29

L.C.M of 17, 23 and 29 = 1 x 17 x 23 x 29

\therefore L.C.M of 17, 23 and 29 = 11339

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

H.C.F of 17, 23 and 29 = 1

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

Hence, H.C.F( 17, 23 and 29) & L.C.M (17, 23 and 2) = 11339.

 

(iii) 8, 9 and 25

The prime factors of 8, 9 and 25 are :  

8 = 2 x 2 x2 = 2³

9 = 3 x 3 = 3²

25 = 5 x 5 = 5²

L.C.M of 8, 9 and 25 = 2³  x 3² x 5² = 2 x 2 x2 × 3 x 3 × 5 x 5

\therefore L.C.M of 8, 9 and 25 = 1800

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

H.C.F of 8, 9 and 25 = 1

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

Hence, H.C.F(8, 9 and 25) = 1  & L.C.M (8, 9 and 25 ) = 1800.

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