Pregunta
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\begin{tabular}{|l|l|}\hline Write 27 as a product of its prime factors: \\ \hline Write multiples of 4 between 10 and 25 \\ \hline Find the HCF and LCM of 24 and 60 \\ \hline\end{tabular}

Ask by Gordon Elliott. in South Africa
Mar 10,2025

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1. **Prime Factors of 27:** \( 27 = 3 \times 3 \times 3 \) or \( 3^3 \). 2. **Multiples of 4 Between 10 and 25:** 12, 16, 20, 24. 3. **HCF and LCM of 24 and 60:** - **HCF:** 12 - **LCM:** 120

Solución

1. **Prime Factorization of 27:** We have \[ 27 = 3 \times 9 = 3 \times 3 \times 3 = 3^3. \] 2. **Multiples of 4 Between 10 and 25:** The multiples of 4 within this range are \[ 12,\ 16,\ 20,\ 24. \] 3. **HCF and LCM of 24 and 60:** First, express each number in its prime factorization: \[ 24 = 2^3 \times 3,\quad 60 = 2^2 \times 3 \times 5. \] - **HCF (Highest Common Factor):** For the common prime factors, take the smallest exponent: \[ \text{HCF} = 2^{\min(3,2)} \times 3^{\min(1,1)} = 2^2 \times 3^1 = 4 \times 3 = 12. \] - **LCM (Least Common Multiple):** For all prime factors, take the largest exponent: \[ \text{LCM} = 2^{\max(3,2)} \times 3^{\max(1,1)} \times 5^{\max(0,1)} = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120. \]

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Mind Expander

To express 27 as a product of its prime factors, we start by recognizing that 27 can be broken down into 3 × 9, and then we further factor 9 into 3 × 3. So, 27 can be represented as \(3^3\), meaning the prime factors are three instances of the prime number 3. The multiples of 4 between 10 and 25 are 12, 16, and 20. These numbers are obtained by taking 4 and multiplying it with whole numbers that fit in that range, specifically 3, 4, and 5 respectively, resulting in these lovely multiples! For the HCF (Highest Common Factor) of 24 and 60, we can break them down into their prime factors: 24 is \(2^3 \times 3\) and 60 is \(2^2 \times 3 \times 5\). The HCF is found by taking the lowest power of common prime factors, resulting in \(2^2 \times 3 = 12\). For the LCM (Lowest Common Multiple), we take the highest power of each prime factor involved, giving us \(2^3 \times 3^1 \times 5^1 = 60\).

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