Problem 6 - Entrance Test

We need to find 3^2023 (mod 7). Let's look at the powers of 3 modulo 7: 3^1 = 3 (mod 7) 3^2 = 9 = 2 (mod 7) 3^3 = 3 * 2 = 6 = -1 (mod 7) 3^4 = 3 * (-1) = -3 = 4 (mod 7) 3^5 = 3 * 4 = 12 = 5 (mod 7) 3^6 = 3 * 5 = 15 = 1 (mod 7) The powers of 3 modulo 7 repeat in a cycle of length 6 (3, 2, 6, 4, 5, 1). To find 3^2023 (mod 7), we need to find the exponent 2023 modulo 6. 2023 / 6 = 337 with a remainder of 1. So, 2023 = 6 * 337 + 1. Therefore, 3^2023 = 3^(6*337 + 1) = (3^6)^337 * 3^1 (mod 7). Since 3^6 = 1 (mod 7): 3^2023 = (1)^337 * 3^1 (mod 7) 3^2023 = 1 * 3 (mod 7) 3^2023 = 3 (mod 7). The remainder when 3^2023 is divided by 7 is 3.