11. Determine the least common multiple of each set of numbers. \( \begin{array}{ll}\text { a) } 20,36,38 & \text { b) } 15,32,44 \\ \text { c) } 12,18,25,30 & \text { d) } 15,20,24,27\end{array} \)
Solución de tutoría real
Responder
Solución
Revisado y aprobado por el equipo de tutoría de UpStudy
Mind Expander
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.