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.