Question

For \( n \in \mathbb{N} \), the largest positive integer that divides \( 81^n + 20n - 1 \) is \( k \). If \( S \) is the sum of all positive divisors of \( k \), then find \( S - k \).

• A. \(117\)
• B. \(130\)
• C. \(115\)
• D. \(127\)

Hint:

Always verify the consistency of \(k\) across multiple values of \(n\) when it must divide a polynomial expression at all \(n\).

Answer 1

The Correct Option is A

To solve the problem, we need to determine the largest positive integer \( k \) that divides \( 81^n + 20n - 1 \) for all \( n \in \mathbb{N} \). Then, we find the sum of all positive divisors of \( k \) and compute \( S - k \). Step 1: Find the Largest Divisor \( k \) We test small values of \( n \) to find a pattern or a common divisor. For \( n = 1 \): \[ 81^1 + 20 \times 1 - 1 = 81 + 20 - 1 = 100 \] For \( n = 2 \): \[ 81^2 + 20 \times 2 - 1 = 6561 + 40 - 1 = 6600 \] For \( n = 3 \): \[ 81^3 + 20 \times 3 - 1 = 531441 + 60 - 1 = 531500 \] We observe that 100 divides all these values. To confirm that 100 is the largest such divisor, we check if 100 divides \( 81^n + 20n - 1 \) for all \( n \). Step 2: Verify Divisibility by 100 We can use modular arithmetic to verify that \( 81^n + 20n - 1 \equiv 0 \pmod{100} \) for all \( n \). First, note that \( 81 \equiv 1 \pmod{20} \), so: \[ 81^n \equiv 1^n \equiv 1 \pmod{20} \] Thus: \[ 81^n + 20n - 1 \equiv 1 + 0 - 1 \equiv 0 \pmod{20} \] Next, check modulo 5: \[ 81 \equiv 1 \pmod{5} \Rightarrow 81^n \equiv 1 \pmod{5} \] \[ 20n \equiv 0 \pmod{5} \] \[ 81^n + 20n - 1 \equiv 1 + 0 - 1 \equiv 0 \pmod{5} \] Since 100 is the least common multiple of 20 and 5, and the expression is divisible by both 20 and 5, it is divisible by 100. Therefore, \( k = 100 \). Step 3: Find the Sum of Divisors of \( k \) The prime factorization of 100 is: \[ 100 = 2^2 \times 5^2 \] The sum of the divisors \( S \) is: \[ S = (1 + 2 + 2^2)(1 + 5 + 5^2) = (1 + 2 + 4)(1 + 5 + 25) = 7 \times 31 = 217 \] Step 4: Compute \( S - k \) \[ S - k = 217 - 100 = 117 \] Final Answer: \[ \boxed{117} \] This corresponds to option (1).