Solution:

Given HCF (Highest Common Factor) = 18 and LCM (Least Common Multiple) = 380 for two numbers.

We need to check if two numbers exist with these given HCF and LCM.

Let the two numbers be a and b.

Prime Factors of 380 = 2^2 × 5 × 19

Prime Factors of 18 = 2 × 3^2

LCM of two numbers = Product of two numbers/HCF of two numbers

Therefore, ab/18 = 380

ab = 18 × 380

ab = 2^2 × 3^2 × 5 × 19 × 2^3 × 5

ab = 2^5 × 3^2 × 5^2 × 19

Now, the HCF of two numbers can only have common factors of the prime factorization of 18.

So, the common factors between a and b are 2 and 3.

Therefore, the two numbers can be written as:

a = 2 × 3 × x

b = 2 × 3 × y

where x and y are co-prime.

ab = 2^5 × 3^2 × 5^2 × 19

(2 × 3 × x) × (2 × 3 × y) = 2^5 × 3^2 × 5^2 × 19

xy = 2^3 × 5 × 19

As x and y are co-prime, they can either be 1 and 1520 or 2 and 760.

Therefore, the two numbers can be:

a = 2 × 3 × 1 = 6

b = 2 × 3 × 1520 = 9120

OR

a = 2 × 3 × 2 = 12

b = 2 × 3 × 760 = 4560

Thus, two numbers can have 18 as their HCF and 380 as their LCM.