To find six distinct numbers such that the GCF of every pair is greater than 1 and the GCF of all six numbers is 1, we can construct the numbers using prime numbers in a systematic way.
Let's consider the following six numbers, each consisting of distinct prime factors:
- \( a = 6 = 2 \times 3 \)
- \( b = 10 = 2 \times 5 \)
- \( c = 14 = 2 \times 7 \)
- \( d = 15 = 3 \times 5 \)
- \( e = 21 = 3 \times 7 \)
- \( f = 35 = 5 \times 7 \)
Check Pairwise GCF
We will check the GCF for each pair of these numbers:
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\( \text{GCF}(a, b) = \text{GCF}(6, 10) = 2 \)
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\( \text{GCF}(a, c) = \text{GCF}(6, 14) = 2 \)
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\( \text{GCF}(a, d) = \text{GCF}(6, 15) = 3 \)
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\( \text{GCF}(a, e) = \text{GCF}(6, 21) = 3 \)
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\( \text{GCF}(a, f) = \text{GCF}(6, 35) = 1 \)
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\( \text{GCF}(b, c) = \text{GCF}(10, 14) = 2 \)
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\( \text{GCF}(b, d) = \text{GCF}(10, 15) = 5 \)
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\( \text{GCF}(b, e) = \text{GCF}(10, 21) = 1 \)
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\( \text{GCF}(b, f) = \text{GCF}(10, 35) = 5 \)
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\( \text{GCF}(c, d) = \text{GCF}(14, 15) = 1 \)
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\( \text{GCF}(c, e) = \text{GCF}(14, 21) = 7 \)
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\( \text{GCF}(c, f) = \text{GCF}(14, 35) = 7 \)
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\( \text{GCF}(d, e) = \text{GCF}(15, 21) = 3 \)
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\( \text{GCF}(d, f) = \text{GCF}(15, 35) = 5 \)
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\( \text{GCF}(e, f) = \text{GCF}(21, 35) = 7 \)
Check GCF of All Six Numbers
Now we need to confirm that the GCF of all six numbers is 1.
The prime factors involved are:
- 2 from 6, 10, and 14
- 3 from 6, 15, and 21
- 5 from 10, 15, and 35
- 7 from 14, 21, and 35
No prime factor can be taken from all six numbers, which means:
\[ \text{GCF}(6, 10, 14, 15, 21, 35) = 1 \]
Conclusion
Thus, the six numbers \( 6, 10, 14, 15, 21, 35 \) meet the requirements:
- The GCF of each pair is greater than 1.
- The GCF of all six numbers is 1.
So, one valid solution is:
Final Set of Numbers: \( 6, 10, 14, 15, 21, 35 \).
1 answer