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    Exercise 10.1.1

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6th class > > Exercise 10.1.1

Exercise 10.1.1

1.Express each number as a product of its prime factors.

(i) 140
Prime factorization of 140 = 2 × × × = 22xx.
(ii) 156
Prime factorization of 156 = 2 × 2 × 3 × 13 = 22 x x .
(iii) 3825
Prime factorization of 3825 = 3 × × 5 × × 17 = 32 x 52 x .
(iv) 5005
Prime factorization of 5005 = 5 × × 11 ×
(v) 7429
Prime factorization of 7429 = × × .

2.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
Prime factors of 26 = 2 × 13 ,91 = 7 × 13,HCF of 26 and 91 = 13
LCM of 26 and 91 = 2 × 7 × 13 = 14 × 13 = .
Product of these two numbers = 26 × 91 = and product of LCM X HCF = 182x13
Thus, the product of two numbers = LCM × HCF
(ii) 510 and 92
Prime factors of 510 = 2 × 3 × 5 × 17,92 = 2 × 2 × 23,HCF of 510 and 92 = 2
LCM of the two numbers = 2 × 2 × 3 × 5 × 17 × 23 = .
Product of these two numbers = 510 × 92 = and Product of LCM x HCF = 2 x 23460 = .
Thus, the product of two numbers = LCM × HCF
(iii) 336 and 54
Prime factors of 336 = 2 × 2 × 2 × 2 × 3 × 7, 54 = 2 × 3 × 3 × 3,HCF of 336 and 54 = 6.
LCM of the two numbers = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = .
Product of these two numbers = 336 × 54= ,Product of LCM x HCF = 3024 × 6 = .
Thus, the product of two numbers = LCM × HCF

3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15 and 21

Prime factors of 12 = × 2 × , 15 = 3 × , 21 = × 7

HCF of 12, 15 and 21 = .

LCM of 12, 15 and 21 = 2² × 3 × 5 × 7 = .

(ii) 17, 23 and 29

Prime factors of 17 = × 1, 23 = × 1, 29 = 29 × 1

HCF of 17, 23 and 29 =

LCM of 17, 23 and 29 = 17 × 23 × 29 =

(iii) 8, 9 and 25

Prime factors of 8 = 2 × × × 1 , 9 = 3 × × 1 ,25 = × × 1.

HCF of 8, 9 and 25 =

LCM of 8 , 9 and 25 = 2 × 2 × 2 × 3 × 3 × 5 × 5 = .

4. Given that HCF (306, 657) = 9, find LCM (306, 657)

Given,HCF (306,657)=9

LCM x 9 = 306 x 657 : LCM = 306x6579

LCM = .

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

If any number ends with the digit 0 that means it should be divisible by 5.

Prime factors of 6n = (2 × 3)n = (2)n × (3)n

Here, 5 is not present in the prime factors of 6n. That means 6n will not be divisible by .

Therefore, 6n cannot end with the digit 0 for any natural number n.

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

So, if the number has more than two factors, it will be composite.

It can be observed that,

7 × 11 × 13 + 13 = 13 (7 × 11 + 1)

= 13 × 78

= 13 × × × × 1

The given number has 2, 3, 13, and 1 as its factors. Therefore, it is a composite number.

Now, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1)

= 5 × (1008 + 1)

= 5 × × 1

Hence, it is a composite number.

7. 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?

Sonia takes 18 minutes per round, and Ravi takes 12 minutes per round.
First, let's find the prime factorizations of 18: x x and 12 : x x .
So,LCM of 18 and 12 is 22x 32 = x = .
Sonia and Ravi will both meet at the starting point again after 36 minutes.