The product of two natural numbers is 15120 and their HCF is 6. Find h...
Problem Statement:
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.
Solution:
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.
Answer:
There are 20 pairs of natural numbers whose product is 15120 and HCF is 6.