What is the remainder when 2^{81} is divided by 17?

By Fermat's Little Theorem, since 17 is prime and gcd(2, 17) = 1, we have 2^16 ≡ 1 (mod 17). Since 81 = 16 × 5 + 1, we get 2^81 = (2^16)^5 × 2^1 ≡ 1^5 × 2 ≡ 2 (mod 17). Wait—let me recalculate: 81 ÷ 16 = 5 remainder 1, so 2^81 ≡ 2^1 ≡ 2 (mod 17)... Actually, checking via modular exponentiation: 2^16 ≡ 1, and 81 = 5(16) + 1, giving 2^81 ≡ 2. But verifying the order: the multiplicative order of 2 modulo 17 is actually 8 (since 2^8 = 256 ≡ 1 mod 17). Since 81 = 10(8) + 1, we have 2^81 ≡ 2^1 ≡ 2 (mod 17)... Re-checking: 81 ÷ 8 = 10 remainder 1. So 2^81 ≡ 2 (mod 17). However, the answer given as D (1) suggests rechecking. Let me verify 2^8 mod 17: 2^4 = 16 ≡ -1, so 2^8 ≡ 1 (mod 17). Then 81 = 10(8) + 1, so 2^81 ≡ 2 (mod 17). The correct answer should be A (2), not D. Given the provided options, option A is correct.

Step-by-step Derivation:
Step 1: Apply Fermat's Little Theorem or find the order of 2 modulo 17.

Step 2: Compute powers of 2 modulo 17:

• 2^1 ≡ 2 (mod 17)
• 2^2 ≡ 4 (mod 17)
• 2^3 ≡ 8 (mod 17)
• 2^4 ≡ 16 ≡ -1 (mod 17)
• 2^8 ≡ (2^4)^2 ≡ (-1)^2 ≡ 1 (mod 17)

Step 3: The multiplicative order of 2 modulo 17 is 8 (the smallest positive integer k such that 2^k ≡ 1 mod 17).

Step 4: Express 81 in terms of 8:
81 = 10 × 8 + 1

Step 5: Calculate 2^81 mod 17:
2^81 = 2^(10×8 + 1) = (2^8)^10 × 2^1 ≡ 1^10 × 2 ≡ 2 (mod 17)

Answer: 2 (Option A)