Question

Can two number have 18 as there hcf and 380. as their lcm 2 give reason

Answer 1

the HCF of two numbers must be a factor of the LCM .
hence the above situation is  possible only if 18 is a factor of 380.

but we seee that it is not a factor of 380.

So it's not possible to have two numbers whose hCF is 18 and LCM is 380.

Alternate method :

if you want the two numbers to be integers,then the answer is NO.

since 18 is their HCF let the two nos. be 18*a and 18*b where a and b be are some positive integers.

Since,

product of LCM and HCF = product of the two numbers

18X380 = 18aX18b

=>18ab = 380

=> 9ab =190

Since 190 is not divisible by 9,a and b be can’t have integer values. So there are no such number.

Or :

Let the two natural numbers be a and b . We know a*b = HFC(a,b) * LCM(a,b) 
since 18 is HCF of a and b we can say a = 18*x , b= 18*y .
18*x*18*y = 18*380
=>18*x*y = 380 
you will find 380 is not divisible by 3 ( so its not divisible by 18) 
and thus we couldn’t find x, y from this. It implies there don’t exist a,b such that they have hcf = 18 and lcm = 380