11. Show that \( 9^{n+1}-8 n-9 \) is divisible by 64 , whenever \( n \) is a positive integer. 12. If \( a \) and \( b \) are distinct integers, prove that \( a-b \) is a factor of \( a^{n}-b^{n} \), whenever \( n \) is positive integer. 13. Show that \( 6^{n+3}-8 n-6 \) is divisible by 6 .

Let's solve each of the problems step by step. ### Problem 11: Show that \( 9^{n+1}-8n-9 \) is divisible by 64 whenever \( n \) is a positive integer. To prove that \( 9^{n+1} - 8n - 9 \) is divisible by 64, we can use mathematical induction. **Base Case:** For \( n = 1 \): \[ 9^{1+1} - 8 \cdot 1 - 9 = 9^2 - 8 - 9 = 81 - 8 - 9 = 64 \] Since 64 is divisible by 64, the base case holds. **Inductive Step:** Assume that for some \( k \geq 1 \), \( 9^{k+1} - 8k - 9 \) is divisible by 64. We need to show that \( 9^{k+2} - 8(k+1) - 9 \) is also divisible by 64. Starting from the inductive hypothesis: \[ 9^{k+2} - 8(k+1) - 9 = 9 \cdot 9^{k+1} - 8k - 8 - 9 \] This can be rewritten as: \[ 9 \cdot 9^{k+1} - 8k - 9 - 8 = 9(9^{k+1} - 8k - 9) + 8(9 - 1) \] Since \( 9^{k+1} - 8k - 9 \) is divisible by 64 (by the inductive hypothesis), we need to check if \( 8(9 - 1) = 8 \cdot 8 = 64 \) is also divisible by 64. Thus, both terms are divisible by 64, and by the principle of mathematical induction, \( 9^{n+1} - 8n - 9 \) is divisible by 64 for all positive integers \( n \). ### Problem 12: If \( a \) and \( b \) are distinct integers, prove that \( a-b \) is a factor of \( a^{n}-b^{n} \), whenever \( n \) is a positive integer. We can use the factor theorem for this proof. The expression \( a^n - b^n \) can be factored as: \[ a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \ldots + b^{n-1}) \] This shows that \( a-b \) is a factor of \( a^n - b^n \). Since \( a \) and \( b \) are distinct integers, \( a-b \neq 0 \). Therefore, \( a-b \) divides \( a^n - b^n \) for any positive integer \( n \). ### Problem 13: Show that \( 6^{n+3}-8n-6 \) is divisible by 6. To prove that \( 6^{n+3} - 8n - 6 \) is divisible by 6, we can analyze the expression modulo 6. **Base Case:** For \( n = 1 \): \[ 6^{1+3} - 8 \cdot 1 - 6 = 6^4 - 8 - 6 = 1296 - 8 - 6 = 1282 \] Calculating \( 1282 \mod 6 \): \[ 1282 \div 6 = 213 \quad \text{remainder } 4 \] So, \( 1282 \equiv 4 \mod 6 \). **Inductive Step:** Assume that for some \( k \), \( 6^{k+3} - 8k - 6 \equiv 0 \mod 6 \). We need to show that \( 6^{k+4} - 8(k+1) - 6 \equiv 0 \mod 6 \). Starting from the inductive hypothesis: \[ 6^{k+4} - 8(k+1) - 6 = 6 \cdot 6^{k+3} - 8k - 8 - 6 \] This can be simplified to: \[ 6(6^{k+3}) - 8k - 14 \] Since \( 6^{k+3} \equiv 0 \mod 6 \), the entire expression is divisible by 6. Thus, by induction, \( 6^{n+3} - 8n - 6 \) is divisible by 6 for all positive integers \( n \). In conclusion: 1. \( 9^{n+1} - 8n - 9 \) is divisible by 64 for positive integers \( n \). 2. \( a-b \) is a factor of \( a^n - b^n \) for distinct integers \( a \) and \( b \). 3. \( 6^{n+3} - 8n - 6 \) is divisible by 6 for positive integers \( n \).