The product of two natural numbers is 15120 and their highest common factor (HCF) is 6. We need to find out how many such pairs exist.



To solve this problem, we will follow the steps below:

1. Prime Factorization of 15120:
- Prime factorize 15120 to find its prime factors:
- 15120 = 2 * 2 * 2 * 2 * 3 * 3 * 3 * 5 * 7
- Write it in exponential form:
- 15120 = 2^4 * 3^3 * 5^1 * 7^1

2. Identify the common factors:
- As the HCF is 6, we need to find the common factors of 6 and 15120.
- The prime factorization of 6 is 2 * 3. So, the common factors are 2 and 3.

3. Generate pairs of factors:
- To find the pairs of natural numbers, we will consider the exponents of the common factors.
- We have 2^4 * 3^3 * 5^1 * 7^1 as the prime factorization of 15120.
- The pairs of exponents for 2 and 3 can be:
- (0, 0), (1, 0), (2, 0), (3, 0), (4, 0) for 2
- (0, 0), (0, 1), (0, 2), (0, 3) for 3

4. Calculate the number of pairs:
- To find the number of pairs, we multiply the number of possibilities for each common factor.
- For factor 2, there are 5 possibilities (0 to 4), and for factor 3, there are 4 possibilities (0 to 3).
- So, the total number of pairs is 5 * 4 = 20.



There are 20 pairs of natural numbers whose product is 15120 and HCF is 6.