To solve for the two numbers given the properties, let's use the relationship between the Highest Common Factor (HCF) and the Lowest Common Multiple (LCM).
Given:
- The HCF of the two numbers is 6.
- The LCM of the two numbers is 60.
We know that:
[tex]\[ \text{Product of the two numbers} = \text{HCF} \times \text{LCM} \][/tex]
Let's denote the two numbers as [tex]\(a\)[/tex] and [tex]\(b\)[/tex].
[tex]\[ a \times b = \text{HCF} \times \text{LCM} \][/tex]
[tex]\[ a \times b = 6 \times 60 \][/tex]
[tex]\[ a \times b = 360 \][/tex]
Given these numbers must be greater than 6 and that their HCF is 6, we should factorize 360 into pairs of factors [tex]\( (a, b) \)[/tex] where the HCF is 6.
So, we need to find pairs such that:
1. [tex]\(a \times b = 360\)[/tex]
2. [tex]\(\text{HCF}(a, b) = 6\)[/tex]
3. [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are both greater than 6.
Let’s explore possible pairs:
1. Start with [tex]\(a = 6\)[/tex].
- [tex]\( b = \frac{360}{6} = 60 \)[/tex]
- Check if HCF(6, 60) is 6.
To check the HCF:
- The prime factorization of 6 is [tex]\(2 \times 3\)[/tex].
- The prime factorization of 60 is [tex]\(2^2 \times 3 \times 5\)[/tex].
The common factors are 2 and 3, which makes the HCF:
[tex]\[ 2 \times 3 = 6 \][/tex]
Therefore, [tex]\(a = 6\)[/tex] and [tex]\(b = 60\)[/tex] correctly satisfy all conditions:
- Both numbers are greater than 6.
- Their HCF is 6.
- Their LCM is 60.
Hence, the two numbers are:
[tex]\[ \boxed{6 \text{ and } 60} \][/tex]
