To find the number of possible pairs of natural numbers that satisfy the given conditions, we need to understand the relationship between the Highest Common Factor (HCF) and the Lowest Common Multiple (LCM) of two numbers.

Let's consider two natural numbers, A and B. The HCF of A and B is the largest number that divides both A and B without leaving a remainder. The LCM of A and B is the smallest multiple that is divisible by both A and B.

The relationship between HCF and LCM can be expressed as follows:
HCF(A, B) * LCM(A, B) = A * B

In this case, we are given that the HCF of two natural numbers is 10 and their LCM is 1000. Using the above relationship, we can write:
10 * 1000 = A * B

Simplifying the equation, we get:
A * B = 10000

Now, we need to find the number of possible pairs of natural numbers (A, B) such that their product is 10000.

To find the factors of 10000, we can start by finding the prime factorization of 10000:
10000 = 2^4 * 5^4

The prime factorization of 10000 can be expressed as the product of its prime factors raised to their respective powers. Since 2 and 5 are prime numbers, the factors of 10000 will be all possible combinations of these prime factors.

For a number to be a factor of 10000, it can take any power of 2 from 0 to 4 (inclusive) and any power of 5 from 0 to 4 (inclusive).

Therefore, the number of possible pairs of natural numbers that satisfy the given conditions is equal to the number of factors of 10000.

To find the number of factors, we add 1 to each of the exponents in the prime factorization and multiply the results:
(4 + 1) * (4 + 1) = 5 * 5 = 25

So, there are 25 possible pairs of natural numbers that satisfy the given conditions.

However, we need to consider that the HCF of the numbers is 10. This means that both A and B must be divisible by 10. Therefore, we need to find the factors of 10000 that are multiples of 10.

The factors of 10000 that are multiples of 10 will have at least one power of 2 and one power of 5.

To find the number of such factors, we subtract the number of factors that have either 0 powers of 2 or 0 powers of 5 from the total number of factors.

The number of factors with 0 powers of 2 is (0 + 1) * (4 + 1) = 5.
The number of factors with 0 powers of 5 is (4 + 1) * (0 + 1) = 5.

Therefore, the number of factors that have either 0 powers of 2 or 0 powers of 5 is 5 + 5 = 10.

Subtracting this from the total number of factors, we get:
25 - 10 = 15

Hence, there are 15 pairs of natural numbers that satisfy the given conditions.

However, we need to consider that the factors must be distinct