Exercise 1.2

Question 1:

Express each number as a product of its prime factors:

i)140

ii)156

iii)3825

iv)5005

v)7429

Solution :

i) 140

Take  LCM of 140 i.e  2×2×5×7×1

Hence,  140=2×2×5×7×1

ii) 156

Take LCM of 156 i.e 2×2×13×3×1

Hence, 156=2×2×13×3×1

iii)3825

Take  LCM of 3825 i.e 3×3×5×5×17×1

Hence, 3825=3×3×5×5×17×1

iv)5005

Take LCM of 5005 i.e 5×7×11×13×1

Hence, 5005=5×7×11×13×1

v)7429

Take LCM of 7429 i.e 17×19×23×1

Hence, 7429=17×19×23×1

Question 2:

Find the LCM and HCF of the following pairs of integer 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

26=2×13×1(expressing as product of it’s prime factors)

91=7×13×1(expressing as product of it’s prime factors)

So, LCM(26,91)=2×7×13×1=182

HCF(26,91)=13×1=13

Verification:

LCM×HCF=13×182=2366

Product of 26 and 91 =2366

Therefore,LCM×HCF=Product of the two numbers .

i) 510 and 92

510=2×3×17×5×1(expressing as product of it’s prime factors)

92=2×2×23×1(expressing as product of it’s prime factors)

So,

LCM(510,92)=2×2×3×5×17×23=23,460

HCF(510,92)=2

Verification:

LCM×HCF=23,460×2=46,920

Product of 510 and 92 =46,920

Therefore,LCM×HCF=Product of the two numbers .

iii) 336 and 54

336=2×2×2×2×7×3×1(expressing as product of it’s prime factors)

54=2×3×3×3×1(expressing as product of it’s prime factors)

So,

LCM(336,54)=24×33×7=3024

HCF(336,54)=2×3=6

Verification:

LCM×HCF=3024×6=18,144

Product of 336 and 54=18,144

Therefore,LCM×HCF=Product of the two numbers .

Question 3:

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

Solution :

i) 12,15 and 21

12=2×2×3

15=5×3

21=7×3

From the above ,HCF(12,15,21)=3and LCM(12,15,21)=420

ii)17,23,and 29

17=17×1

23=23×1

29=29×1

From the above ,HCF(17,23,29)=1and LCM(17,23,29)=11339

iii)8,9 and 25

8=2×2×2

9=3×3

25=5×5

From the above ,HCF(8,9,25)=1and LCM(8,9,25)=1800

Question 4:

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

Solution :

We know,

HCF×LCM=Product of two numbers

i.e 9×LCM=306×657

LCM=306×6579=22338

Question 5:

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

Solution :

If the number 6n ends with the digit zero,then it is divisible by 5.Therefore the prime factorization of 6n contains the prime 5.This is not possible because the only prime in the factorization of 6n is 2 and 3 and the uniqueness of the fundamental theorem of arithmetic guarantees that these are no other prime in the factorization of 6n

Hence, it is very clear that there is no value of n in natural number for which 6n ends with the digit zero.

Question 6:

Explain why 7×11×13+13 and 7×6×5×4×3×2×1+5 are composite number.

Solution :

We have,

7×11×13+13

=13(7×11×1+1)

=13×78

=13×3×2×13

Hence, it is a composite number .

We have,               7×6×5×4×3×2×1+5

=5(7×6×4×3×2×1+1)

=5(1008+1)

=5×1009

Hence, it is a composite number .

Question 6:

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?

Solution :

Since, Both Sonia and Ravi move in the same direction and at the same time, the method to find the time when they will be meeting again at the starting point is LCM(18,12) is 2×3×3×2×1=36

Therefore, Sonia and Ravi will meet again after 36 minutes.


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Exercise 1.1