Exercise 3.4

1. Find the HCF of the following numbers by prime factorisation and continued division method?

(i) 18, 27, 36 (ii) 106, 159, 265

(iii) 10, 35, 40 (iv) 32, 64, 96, 128

(i) Solution

18=2×3×3=2×32

27=3×3×3=33

36=2×2×3×3=22×32

The common factor among 18, 27, and 36 is 32 = .

Thus, HCF = .

HCF of 18 and 27

Divide 27 by 18 = remainder

Divide 18 by 9 = remainder

HCF(18, 27) =

HCF of 9 and 36

Divide 36 by 9 = remainder

HCF(9, 36) =

Thus, HCF(18, 27, 36) = .

(ii) 106, 159, 265

106=2×53

159=3×53

265=5×53

The common factor among 106, 159, and 265 is 53.

Thus, HCF =

HCF of 106 and 159

Divide 159 by 106 = remainder

Divide 106 by 53 = remainder

HCF(106, 159) =

HCF of 53 and 265

Divide 265 by 53 = remainder

HCF(53, 265) =

Thus, HCF(106, 159, 265) =

(iii) 10, 35, 40

10=2×5

35=5×7

40=23×5

The common factor among 10, 35, and 40 is 5.

Thus, HCF =

HCF of 10 and 35

Divide 35 by 10 = remainder

Divide 10 by 5 = remainder

HCF(10, 35) =

HCF of 5 and 40

Divide 40 by 5 = remainder

HCF(5, 40) =

Thus, HCF(10, 35, 40) =

(iv) 32, 64, 96, 128

32=25

64=26

96=25×3

128=27

The common factor among 32, 64, 96, and 128 is 25 = 32.

Thus, HCF =

HCF of 32 and 64

Divide 64 by 32 = remainder

HCF(32, 64) =

HCF of 32 and 96

Divide 96 by 32 = remainder

HCF(32, 96) =

HCF of 32 and 128

Divide 128 by 32 = remainder

HCF(32, 128) =

Thus, HCF(32, 64, 96, 128) =

2. Find the largest number which is a factor of each of the numbers 504, 792 and 1080?

First, let's find the prime factorization of each number:

504 = 2³3³ × 3²2² ×

792 = 2³4³ × 3²2² ×

1080 = 2³1³ × 3³1³ ×

For the GCD, we take each prime factor to the lowesthighest power that appears in any of the numbers:

Looking at each prime factor: 2 appears with power in all numbers.

3 appears with power in two numbers and power in one number.

5, 7, and 11 each appear in only one number, so we don't include them.

Therefore, GCD = 2³3³ × 3²5² = × =

To verify: 504 ÷ 72 = (whole number)

792 ÷ 72 = (whole number)

1080 ÷ 72 = (whole number)

Thus, 72 is the largest number which is a factor of all three numbers.

3. The length, breadth and height of a room are 12m, 15m and 18m respectively. Determine the length of longest stick that can measure all the dimensions of the room in exact number of times ?

HCF of 12 , 15 , 18 will be the longest tape measure.

12 = × = 3 × ×

15 = ×

18 = 9 × = × ×

Therefore HCF of 12 , 15 , 18 is

Therefore, m is the longest tape measure which can measure all the dimensions.

4. HCF of co-prime numbers 4 and 15 was found as follows by factorisation: 4 = 2 x 2 and 15 = 3 x 5 Since there is no common prime factor, HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?

We have been given that,

4 = × and 15 = × .

Since there are no common prime factors, so HCF of 4 and 15 is 0 is incorrect.

We see that there are no prime numbers or prime factors common to 4 and 15. Therefore their HCF is the universal factor .

Hence, 4 and 15 are co-prime numbers having HCF as and not .

5. What is the capacity of the largest vessel which can empty the oil from three vessels containing 32 litres, 24 litres and 48 litres an exact number of times?

By Prime factorisation

factors of 32 = × × × ×

factors of 24 = × × ×

factors of 48 = × × × ×

HCF = × × =

Therefore, the capacity of the largest vessel which can empty the oil from three containers is .