## S. K. Gupta Anubhuti Gangal Class 5 Math Fourth Chapter Factors And Multiples Exercise 13

## EXERCISE 13

**(1) Write all the factors of each of the following numbers and then tell whether it is a prime number or a composite number.**

(a) 11 = 1, 11 (Prime)

(b) 32 = 1, 2, 4, 8, 16, 32 (Composite)

(c) 40 = 1, 2, 4, 5, 8, 10, 20, 40 (Composite)

(d) 19 = 1, 19 (Prime)

(e) 18 = 1, 2, 3, 6, 9, 18 (Composite)

(f) 41 = 1, 41 (Prime)

**(2) Complete the factor tree. Then write the prime factorization.**

**(3) Find the prime factorization of each of the following numbers.**

(a) 216 = 2 × 2 × 2 × 3 × 3 × 3

(b) 150 = 2 × 3 × 5 × 5

(c) 84 = 2 × 2 × 3 ×7

(d) 1080 = 2 × 2 × 2 × 3 × 3 × 3 × 5

**(4) List the common factors of each of the following sets of numbers. Hence find their HCF.**

(a) 15, 35

Solution: Factors of 15 = 1, 3, 5

Factors of 35 = 1, 5, 7

The common factors are 1, 5

Out of these the common factor, 5 is largest.

∴ Therefore the HCF of 15 and 35 is 5.

(b) 16, 27

Solution: Factors of 16 = 1, 2, 4, 8

Factors of 27 = 1, 3, 9

The common factors are 1

Out of these the common factor, 1 is largest.

∴ Therefore the HCF of 16 and 27 is 1.

(c) 30, 42

Solution: Factors of 30 = 1, 2, 3, 5, 6, 10, 15

Factors of 42 = 1, 2, 3, 6, 8, 14, 21

The common factors are 1, 2, 3, 6

Out of these the common factor, 6 is largest.

∴ Therefore the HCF of 30 and 42 is 6.

(d) 18, 32, 48

Solution: Factors of 18 = 1, 2, 3, 6, 9

Factors of 32 = 1, 2, 4, 8, 16

Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24

The common factors are 1, 2

Out of these the common factor, 2 is largest.

∴ Therefore the HCF of 18, 32 and 48 is 2.

**(5) Find the HCF of the following by prime factorization method.**

(a) 36, 60

Solution: All the factors of 36 are 2 × 2 × 3 × 3

All the factors of 60 are 2 × 2 × 3 × 5

Common factors of 36 and 60 are 2 × 2 × 3 = 12

Thus, the highest common factor of 36 and 60 is 12

Hence, HCF of 36 and 60 = 12

(b) 18, 24, 42

Solution: All the factors of 18 are 2 × 3 × 3

All the factors of 24 are 2 × 2 × 2 × 3

All the factors are of 42 are 2 × 3 × 7

Common factors of 18, 24 and 42 are 2 × 3 = 6

Thus, the highest common factor of 18, 24 and 42 is 6.

(c) 36, 48, 60

Solution: All the factors of 36 are 2 × 2 × 3 × 3

All the factors of 48 are 2 × 2 × 2 ×2 × 3

All the factors are of 60 are 2 × 2 × 3 × 5

Common factors of 36, 48 and 60 are 2 × 2 × 3 = 12.

Thus, the highest common factor of 36, 48 and 60 is 12.

(d) 28, 42, 56

Solution: All the factors of 28 are 2 × 2 × 7

All the factors of 42 are 2 × 3 × 7

All the factors are of 56 are 2 × 2 × 2 × 7

Common factors of 28, 42 and 56 are 2 × 7 = 14.

Thus, the highest common factor of 28, 42 and 56 is 14.

**(6) Find the HCF of the following by long division method.**

(a) 128 and 192

Solution:

Hence, HCF of 128 and 192 is

(b) 145 and 325

Solution:

Hence, HCF of 145 and 325 is

(c) 495 and 945

Solution:

Hence, HCF of 495 and 945 is

(7) Find the LCM of the following by prime factorization method.

(a) 24 and 32

Solution: 24 = 2 × 2 × 2 × 3

32 = 2 × 2 × 2 × 2 × 2

LCM = 2 × 2 × 2 × 2 × 2 × 3 = 96

(b) 56 and 32

Solution: 56 = 2 × 2 × 2 × 7

32 = 2 × 2 × 2 × 2 × 2

LCM = 2 × 2 × 2 × 2 × 2 × 7 = 224.

(c) 25 and 40

Solution: 25 = 5 × 5

40 = 2 × 2 × 2 × 5

LCM = 2 × 2 × 2 × 5 × 5 = 200.

**(8) Find the LCM of the following by division method.**

(a) 4, 12 and 20

∴ LCM = 2 × 2 × 1 × 3 × 5 = 60.

(b) 16, 28, 44

∴ LCM = 2 × 2 × 4 × 7 ×11 = 1232.

(c) 39, 45, 54

∴ LCM = 3 × 3 × 13 × 5 × 6 = 3510.

(d) 25, 40, 80, 120

∴ LCM = 2 × 2 × 2 × 5 × 5 × 1 × 2 × 3 = 1200.

I am happy to get solution