Pregunta
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11. Determine the least common multiple of each set of numbers. \( \begin{array}{ll}\text { a) } 20,36,38 & \text { b) } 15,32,44 \\ \text { c) } 12,18,25,30 & \text { d) } 15,20,24,27\end{array} \)

Ask by Mann Fuentes. in Canada
Feb 03,2025

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The LCMs are: - a) 3,420 - b) 5,280 - c) 900 - d) 1,080

Solución

To determine the **least common multiple (LCM)** of a set of numbers, follow these steps: 1. **Prime Factorization**: Break down each number into its prime factors. 2. **Identify Highest Powers**: For each prime number that appears in the factorizations, take the highest power of that prime. 3. **Multiply Together**: Multiply these highest powers to get the LCM. Let's apply this method to each part of the problem: --- ### **a) LCM of 20, 36, 38** 1. **Prime Factorization**: - \(20 = 2^2 \times 5\) - \(36 = 2^2 \times 3^2\) - \(38 = 2 \times 19\) 2. **Identify Highest Powers**: - \(2^2\) (from 20 and 36) - \(3^2\) (from 36) - \(5\) (from 20) - \(19\) (from 38) 3. **Multiply Together**: \[ \text{LCM} = 2^2 \times 3^2 \times 5 \times 19 = 4 \times 9 \times 5 \times 19 = 3420 \] **Answer:** The LCM of 20, 36, and 38 is **3,420**. --- ### **b) LCM of 15, 32, 44** 1. **Prime Factorization**: - \(15 = 3 \times 5\) - \(32 = 2^5\) - \(44 = 2^2 \times 11\) 2. **Identify Highest Powers**: - \(2^5\) (from 32) - \(3\) (from 15) - \(5\) (from 15) - \(11\) (from 44) 3. **Multiply Together**: \[ \text{LCM} = 2^5 \times 3 \times 5 \times 11 = 32 \times 3 \times 5 \times 11 = 5,280 \] **Answer:** The LCM of 15, 32, and 44 is **5,280**. --- ### **c) LCM of 12, 18, 25, 30** 1. **Prime Factorization**: - \(12 = 2^2 \times 3\) - \(18 = 2 \times 3^2\) - \(25 = 5^2\) - \(30 = 2 \times 3 \times 5\) 2. **Identify Highest Powers**: - \(2^2\) (from 12) - \(3^2\) (from 18) - \(5^2\) (from 25) 3. **Multiply Together**: \[ \text{LCM} = 2^2 \times 3^2 \times 5^2 = 4 \times 9 \times 25 = 900 \] **Answer:** The LCM of 12, 18, 25, and 30 is **900**. --- ### **d) LCM of 15, 20, 24, 27** 1. **Prime Factorization**: - \(15 = 3 \times 5\) - \(20 = 2^2 \times 5\) - \(24 = 2^3 \times 3\) - \(27 = 3^3\) 2. **Identify Highest Powers**: - \(2^3\) (from 24) - \(3^3\) (from 27) - \(5\) (from 15 and 20) 3. **Multiply Together**: \[ \text{LCM} = 2^3 \times 3^3 \times 5 = 8 \times 27 \times 5 = 1,080 \] **Answer:** The LCM of 15, 20, 24, and 27 is **1,080**. --- ### **Summary of Answers** - **a)** 3,420 - **b)** 5,280 - **c)** 900 - **d)** 1,080

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To find the least common multiple (LCM) of a set of numbers, you can use their prime factorization. For the numbers \(20, 36, 38\): - \(20 = 2^2 \times 5\) - \(36 = 2^2 \times 3^2\) - \(38 = 2^1 \times 19\) Take the highest power of each prime factor: - \(2^2\) from \(20\) or \(36\) - \(3^2\) from \(36\) - \(5^1\) from \(20\) - \(19^1\) from \(38\) Multiply these together: \( LCM = 2^2 \times 3^2 \times 5^1 \times 19^1 = 4 \times 9 \times 5 \times 19 = 3420 \). Now for the numbers \(15, 32, 44\): - \(15 = 3^1 \times 5^1\) - \(32 = 2^5\) - \(44 = 2^2 \times 11\) Taking the highest power of each prime: - \(2^5\) from \(32\) - \(3^1\) from \(15\) - \(5^1\) from \(15\) - \(11^1\) from \(44\) Calculate: \( LCM = 2^5 \times 3^1 \times 5^1 \times 11^1 = 32 \times 3 \times 5 \times 11 = 5280 \). Now for the numbers \(12, 18, 25, 30\): - \(12 = 2^2 \times 3^1\) - \(18 = 2^1 \times 3^2\) - \(25 = 5^2\) - \(30 = 2^1 \times 3^1 \times 5^1\) Taking the highest power of each prime: - \(2^2\) - \(3^2\) - \(5^2\) Calculate: \( LCM = 2^2 \times 3^2 \times 5^2 = 4 \times 9 \times 25 = 900 \). Lastly, for the numbers \(15, 20, 24, 27\): - \(15 = 3^1 \times 5^1\) - \(20 = 2^2 \times 5^1\) - \(24 = 2^3 \times 3^1\) - \(27 = 3^3\) Taking the highest power of each prime: - \(2^3\) from \(24\) - \(3^3\) from \(27\) - \(5^1\) from \(15\) Calculate: \( LCM = 2^3 \times 3^3 \times 5^1 = 8 \times 27 \times 5 = 1080 \). So, the final results are: a) 3420, b) 5280, c) 900, d) 1080.

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