- Use the intersection-of-sets method: list all divisors to find the largest common one (GCD) and list multiples to find the smallest common one (LCM).
- The GCD and LCM for the given sets are:
- a. \(\operatorname{GCD}(50, 20) = \boxed{10}\), \(\operatorname{LCM}(50, 20) = \boxed{100}\)
- b. \(\operatorname{GCD}(28, 36) = \boxed{4}\), \(\operatorname{LCM}(28, 36) = \boxed{252}\)
- c. \(\operatorname{GCD}(8, 88, 64) = \boxed{8}\), \(\operatorname{LCM}(8, 88, 64) = \boxed{704}\)
- d. \(\operatorname{GCD}(11, 9) = \boxed{1}\), \(\operatorname{LCM}(11, 9) = \boxed{99}\)
Explanation
- GCD and LCM of 50 and 20
To find the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) using the intersection-of-sets method, we first list all divisors for each number and then all multiples for each number. The GCD is the largest value in the intersection of the sets of divisors, while the LCM is the smallest positive value in the intersection of the sets of multiples.
- Divisors of 50: \(D_{50} = \{1, 2, 5, 10, 25, 50\}\)
- Divisors of 20: \(D_{20} = \{1, 2, 4, 5, 10, 20\}\)
- Intersection of divisors: \(D_{50} \cap D_{20} = \{1, 2, 5, 10\}\)
- The greatest element is 10, so \(\operatorname{GCD}(50, 20) = 10\).
- Multiples of 50: \(M_{50} = \{50, 100, 150, 200, \dots\}\)
- Multiples of 20: \(M_{20} = \{20, 40, 60, 80, 100, 120, \dots\}\)
- The smallest common multiple is 100, so \(\operatorname{LCM}(50, 20) = 100\).
- GCD and LCM of 28 and 36
- Divisors of 28: \(D_{28} = \{1, 2, 4, 7, 14, 28\}\)
- Divisors of 36: \(D_{36} = \{1, 2, 3, 4, 6, 9, 12, 18, 36\}\)
- Intersection of divisors: \(D_{28} \cap D_{36} = \{1, 2, 4\}\)
- The greatest element is 4, so \(\operatorname{GCD}(28, 36) = 4\).
- Multiples of 28: \(M_{28} = \{28, 56, 84, \dots, 224, 252, \dots\}\)
- Multiples of 36: \(M_{36} = \{36, 72, 108, \dots, 216, 252, \dots\}\)
- The smallest common multiple is 252, so \(\operatorname{LCM}(28, 36) = 252\).
- GCD and LCM of 8, 88, and 64
- Divisors of 8: \(D_{8} = \{1, 2, 4, 8\}\)
- Divisors of 88: \(D_{88} = \{1, 2, 4, 8, 11, 22, 44, 88\}\)
- Divisors of 64: \(D_{64} = \{1, 2, 4, 8, 16, 32, 64\}\)
- Intersection of divisors: \(D_{8} \cap D_{88} \cap D_{64} = \{1, 2, 4, 8\}\)
- The greatest element is 8, so \(\operatorname{GCD}(8, 88, 64) = 8\).
- Multiples of 8: \(M_{8} = \{8, 16, \dots, 640, 648, 656, 664, 672, 680, 688, 696, 704, \dots\}\)
- Multiples of 88: \(M_{88} = \{88, 176, \dots, 616, 704, \dots\}\)
- Multiples of 64: \(M_{64} = \{64, 128, \dots, 640, 704, \dots\}\)
- The smallest common multiple is 704, so \(\operatorname{LCM}(8, 88, 64) = 704\).
- GCD and LCM of 11 and 9
- Divisors of 11: \(D_{11} = \{1, 11\}\)
- Divisors of 9: \(D_{9} = \{1, 3, 9\}\)
- Intersection of divisors: \(D_{11} \cap D_{9} = \{1\}\)
- The greatest element is 1, so \(\operatorname{GCD}(11, 9) = 1\).
- Multiples of 11: \(M_{11} = \{11, 22, 33, 44, 55, 66, 77, 88, 99, \dots\}\)
- Multiples of 9: \(M_{9} = \{9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, \dots\}\)
- The smallest common multiple is 99, so \(\operatorname{LCM}(11, 9) = 99\).
- Final Answer Summary
The results for each pair/set are:
A\(\operatorname{GCD}(50, 20) = 10\), \(\operatorname{LCM}(50, 20) = 100\)
B\(\operatorname{GCD}(28, 36) = 4\), \(\operatorname{LCM}(28, 36) = 252\)
C\(\operatorname{GCD}(8, 88, 64) = 8\), \(\operatorname{LCM}(8, 88, 64) = 704\)
D\(\operatorname{GCD}(11, 9) = 1\), \(\operatorname{LCM}(11, 9) = 99\)
Examples
Imagine you are planning to organize items in a warehouse. You have two types of shipping boxes that are 50 cm and 20 cm wide. If you want to line them up in separate rows so they occupy the same total length, you are looking for the Least Common Multiple (LCM), which in this case is 100 cm. Similarly, if you want to divide a large area into smaller, equal-sized square tiles that fit perfectly into dimensions like 50 cm and 20 cm, the largest tile size you can use is the Greatest Common Divisor (GCD), which is 10 cm. This helps in efficient space planning and resource management!