b Use your answers to part a to find the smallest positive integer value of \( n \) for which \( 225 n \) is a multiple of 540 .

To find the smallest positive integer value of \( n \) such that \( 225n \) is a multiple of \( 540 \), we first need to determine the prime factorizations of \( 225 \) and \( 540 \). 1. **Factorization of \( 225 \)**: \[ 225 = 15^2 = (3 \cdot 5)^2 = 3^2 \cdot 5^2 \] 2. **Factorization of \( 540 \)**: \[ 540 = 54 \cdot 10 = (2 \cdot 27) \cdot (2 \cdot 5) = 2^2 \cdot 3^3 \cdot 5 \] Now we have: - \( 225 = 3^2 \cdot 5^2 \) - \( 540 = 2^2 \cdot 3^3 \cdot 5^1 \) Next, we want \( 225n \) to be a multiple of \( 540 \). This means that the prime factorization of \( 225n \) must include at least the prime factors of \( 540 \) with the same or higher powers. The prime factorization of \( 225n \) is: \[ 225n = 3^2 \cdot 5^2 \cdot n \] To ensure \( 225n \) is a multiple of \( 540 \), we need to satisfy the following conditions for each prime factor: - For \( 2 \): \( 540 \) has \( 2^2 \), but \( 225 \) has no factor of \( 2 \). Therefore, \( n \) must provide at least \( 2^2 \). - For \( 3 \): \( 540 \) has \( 3^3 \) and \( 225 \) has \( 3^2 \). Thus, \( n \) must provide at least \( 3^{3-2} = 3^1 \). - For \( 5 \): \( 540 \) has \( 5^1 \) and \( 225 \) has \( 5^2 \). Since \( 225 \) already has enough \( 5 \)'s, \( n \) does not need to provide any additional factors of \( 5 \). Now we can summarize the requirements for \( n \): \[ n \text{ must include at least } 2^2 \text{ and } 3^1. \] Thus, the smallest \( n \) that satisfies these conditions is: \[ n = 2^2 \cdot 3^1 = 4 \cdot 3 = 12. \] Therefore, the smallest positive integer value of \( n \) for which \( 225n \) is a multiple of \( 540 \) is: \[ \boxed{12}. \]