# Find the LCM and HCF of the following integers by applying the prime factorization method .

(i) 12, 15 and 21

12 = 2 x 3

15 = 3 x 5

21 = 3 x 7

HCF = 3

LCM = 2 x 3 x 5 x 7 = 42O

II) 17, 23 and 29

17 = 1 x 17

23 = 1 x 23

29 = 1 x 29

HCF = 1

LCM = 17 x 23 x 29 = 11339

III) 8, 9 and 25

8 = 2 x 2 x 2

9 = 3 x 3

25 = 5 x 5

HCF = 1

LCM = 2 x 2 x 2 x 3 x 3 x 5 x 5 = 1800

Ques – Given that HCF (306, 657) = 9, find LCM (306, 657)

Ans. HCF (306, 657) =9

We know that LCM x HCF = Product of two numbers

Since LCM x HCF = 306 x 657

LCM = 306 x 657/ HCF = 306 x 657/ 9

LCM = 22338

Ques – check whether 6n can end with the digit 0 for any natural number n.

Ans. if any number end with the digit O, it should be divisible by 10 or in other word, it will also be divisible by 2 and 5 as 1O = 2 x 5

Prime factorisation of 6n = (2 x 3)n

It can be observed that 5 is not in the prime factorisation of 6n.

Hence, for any value of n, 6n will not be divisible by 5.

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

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

Ans. numbers are of two types – prime and composite. Prime numbers can be divided by 1 and 0nly itself, whereas c0mp0site numbers have factors 0ther than 1 and itself.

It can be 0bseved that

7 x 11 x 13 + 13 = 13 x ( 7 x 11 + 1) = 13 x (77 + 1)

= 13 x 78

= 13 x 13 x 6

The given expressi0n has 6 and 13 as its fact0rs. Theref0re, it is a c0mp0site number.

7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x (7 x 6 x 4 x 3 x 2 x 1 + 1)

= 5 x (1008 + 1)

= 5 x 1009

1009 as its fact0rised father. Therefore, the given expression has 5 and 1009 as its fact0rs. Hence, it is a c0mp0site number.