(i) 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7
(ii) 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13
(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17
(iv) 5005 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23

(i) 26 = 2 × 13
91 =7 × 13
HCF = 13
LCM =2 × 7 × 13 =182
Product of two numbers 26 × 91 = 2366
Product of HCF and LCM 13 × 182 = 2366
Hence, product of two numbers = product of HCF × LCM

(ii) 510 = 2 × 3 × 5 × 17
92 =2 × 2 × 23
HCF = 2
LCM =2 × 2 × 3 × 5 × 17 × 23 = 23460
Product of two numbers 510 × 92 = 46920
Product of HCF and LCM 2 × 23460 = 46920
Hence, product of two numbers = product of HCF × LCM

(iii) 336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
HCF = 2 × 3 = 6
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 =3024
Product of two numbers 336 × 54 =18144
Product of HCF and LCM 6 × 3024 = 18144
Hence, product of two numbers = product of HCF × LCM

(i) 12 = 2 × 2 × 3
15 =3 × 5
21 =3 × 7
HCF = 3
LCM = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
HCF = 1
LCM = 1 × 17 × 19 × 23 = 11339

(iii) 8 =1 × 2 × 2 × 2
9 =1 × 3 × 3
25 =1 × 5 × 5
HCF =1
LCM = 1 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

We have the formula that
Product of LCM and HCF = product of number
LCM × 9 = 306 × 657
Divide both side by 9 we get
LCM = (306 × 657) / 9 = 22338

If any digit has last digit 10 that means it is divisible by 10 and the factors of 10 = 2 × 5.

So value 6n should be divisible by 2 and 5 both 6n is divisible by 2 but not divisible by 5 So it can not end with 0.

7 × 11 × 13 + 13
Taking 13 common, we get
13 (7 x 11 +1 )
13(77 + 1 )
13 (78)
It is product of two numbers and both numbers are more than 1 so it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
Taking 5 common, we get
5(7 × 6 × 4 × 3 × 2 × 1 +1)
5(1008 + 1)
5(1009)
It is product of two numbers and both numbers are more than 1 so it is a composite number

They will be meet again after LCM of both values at the starting point.
18 = 2 × 3 × 3
12 = 2 × 2 × 3
LCM = 2 × 2 × 3 × 3 = 36
Therefore, they will meet together at the starting point after 36 minutes.

Let take √5 as rational number
If a and b are two co prime number and b is not equal to 0.
We can write √5 = a/b
Multiply by b both side we get
b√5 = a
To remove root, Squaring on both sides, we get
5b² = a² …  (i) 
 

Therefore, 5 divides a² and according to theorem of rational number, for any prime number pwhich is divides a² then it will divide a also.
That means 5 will divide a. So we can write
a = 5c
Putting value of a in equation (i) we get
5b² = (5c)²
5b² = 25c²
Divide by 25 we get

b2/5 = c²

Similarly, we get that b will divide by 5
and we have already get that a is divide by 5
but a and b are co prime number. so it contradicts.
Hence √5 is not a rational number, it is irrational.