Question

Find three different numbers such that the GCF of each pair of these numbers is greater than 1 and the GCF of all the numbers is 1.

Answer 1

To find three different numbers such that the GCF of each pair is greater than 1 but the GCF of all three numbers is 1, we can look for numbers that share common prime factors in pairs but have distinct prime factors across all three.

A good example of such numbers is:
- \(6\) (which has the prime factors \(2\) and \(3\))
- \(10\) (which has the prime factors \(2\) and \(5\))
- \(15\) (which has the prime factors \(3\) and \(5\))

Now let's calculate the GCF for each pair:

1. **GCF(6, 10)**:
- The prime factors are \(2, 3\) (for 6) and \(2, 5\) (for 10).
- Common factor: \(2\). Thus, \( \text{GCF}(6, 10) = 2 \).

2. **GCF(6, 15)**:
- The prime factors are \(2, 3\) (for 6) and \(3, 5\) (for 15).
- Common factor: \(3\). Thus, \( \text{GCF}(6, 15) = 3 \).

3. **GCF(10, 15)**:
- The prime factors are \(2, 5\) (for 10) and \(3, 5\) (for 15).
- Common factor: \(5\). Thus, \( \text{GCF}(10, 15) = 5 \).

Now we check the GCF of all three numbers:

- **GCF(6, 10, 15)**:
- The prime factors are \(2, 3\) (for 6), \(2, 5\) (for 10), and \(3, 5\) (for 15).
- There are no common prime factors among all three numbers. Therefore, \( \text{GCF}(6, 10, 15) = 1 \).

Hence, the three numbers \(6\), \(10\), and \(15\) satisfy all the conditions:
- GCF of each pair is greater than 1.
- GCF of all three numbers is 1.

Final answer: \(6\), \(10\), \(15\).