1. **Problem Statement:** We are working with a multiplication operation \(\otimes\) modulo 8 on the set \(\{3,5,7\}\). 2. **Step (a): Draw the multiplication table \(\otimes\) modulo 8 for \(\{3,5,7\}\):** Calculate each product modulo 8: - \(3 \otimes 3 = (3 \times 3) \bmod 8 = 9 \bmod 8 = 1\) - \(3 \otimes 5 = (3 \times 5) \bmod 8 = 15 \bmod 8 = 7\) - \(3 \otimes 7 = (3 \times 7) \bmod 8 = 21 \bmod 8 = 5\) - \(5 \otimes 3 = (5 \times 3) \bmod 8 = 15 \bmod 8 = 7\) - \(5 \otimes 5 = (5 \times 5) \bmod 8 = 25 \bmod 8 = 1\) - \(5 \otimes 7 = (5 \times 7) \bmod 8 = 35 \bmod 8 = 3\) - \(7 \otimes 3 = (7 \times 3) \bmod 8 = 21 \bmod 8 = 5\) - \(7 \otimes 5 = (7 \times 5) \bmod 8 = 35 \bmod 8 = 3\) - \(7 \otimes 7 = (7 \times 7) \bmod 8 = 49 \bmod 8 = 1\) The table is: \[ \begin{array}{c|ccc} \otimes & 3 & 5 & 7 \\ \hline 3 & 1 & 7 & 5 \\ 5 & 7 & 1 & 3 \\ 7 & 5 & 3 & 1 \end{array} \] 3. **Step (b)(i): Evaluate \((7 \otimes 3) \otimes 5\) and \(7 \otimes (3 \otimes 5)\):** - From the table, \(7 \otimes 3 = 5\) - Then, \((7 \otimes 3) \otimes 5 = 5 \otimes 5 = 1\) - Also, \(3 \otimes 5 = 7\) - Then, \(7 \otimes (3 \otimes 5) = 7 \otimes 7 = 1\) Both expressions equal \(1\). 4. **Step (b)(ii): Find the truth set of \(n \otimes (n+2) = 3\) for \(n \in \{3,5,7\}\):** Calculate \(n \otimes (n+2)\) modulo 8: - For \(n=3\), \(3+2=5\), \(3 \otimes 5 = 7 \neq 3\) - For \(n=5\), \(5+2=7\), \(5 \otimes 7 = 3\) (true) - For \(n=7\), \(7+2=9 \equiv 1 \bmod 8\), but 1 is not in the set, so no value for \(7 \otimes 1\) in the table. Since \(7+2\) modulo 8 is 1, which is not in the set, we cannot evaluate \(7 \otimes 1\) here. Therefore, the truth set is \(\{5\}\). **Final answers:** - (b)(i) \((7 \otimes 3) \otimes 5 = 1\) and \(7 \otimes (3 \otimes 5) = 1\) - (b)(ii) Truth set of \(n \otimes (n+2) = 3\) is \(\{5\}\).