The minimum positive integer p such that 3p modulo 17 = 1 is

Key idea: by Fermat's little theorem, 3^16 ≡ 1 (mod 17), so the multiplicative order of 3 modulo 17 must divide 16.

1. Check the proper divisors of 16: 1, 2, 4, 8.

2. Compute powers modulo 17: 3^1 ≡ 3; 3^2 ≡ 9; 3^4 = (3^2)^2 ≡ 9^2 = 81 ≡ 13 (mod 17); 3^8 = (3^4)^2 ≡ 13^2 = 169 ≡ 16 (mod 17). None of these are 1.

3. Since no smaller divisor of 16 gives 3^d ≡ 1, the order of 3 modulo 17 is 16, so the minimal positive p with 3^p ≡ 1 (mod 17) is p = 16.

Answer: 16