Discussion

Answer: A

This requires Wilson's Theorem, which states that for a prime number p, \((p-1)! \equiv -1 \pmod{p}\). Here, p=23. So, \(22! \equiv -1 \pmod{23}\). We can write \(22! = 22 \times 21 \times 20!\). Since \(22 \equiv -1 \pmod{23}\) and \(21 \equiv -2 \pmod{23}\), the equation becomes \((-1) \times (-2) \times (20!) \equiv -1 \pmod{23}\). This simplifies to \(2 \times (20!) \equiv -1 \equiv 22 \pmod{23}\). Dividing by 2, we get \(20! \equiv 11 \pmod{23}\). Wait, let me re-check. Is there a simpler way? Let's check my Wilson's theorem application. \((p-2)! \equiv 1 \pmod p\). So \(21! \equiv 1 \pmod{23}\). Now \(21! = 21 \times 20! \equiv -2 \times 20! \pmod{23}\). So \(-2 \times 20! \equiv 1 \pmod{23}\). We need to find the inverse of -2 mod 23. \(-2x \equiv 1\). \(2x \equiv -1 \equiv 22\). \(x=11\). I'm still getting 11. Let me re-read the theorem again. Yes, \((p-2)! \equiv 1 \pmod p\). Okay, maybe the question has a simpler pattern. Let's try \((16!)\%17\). No, \((15!)\%17\). \(16! \equiv -1\). \(16 \times 15! \equiv -1\). \(-1 \times 15! \equiv -1\). So \(15! \equiv 1 \pmod{17}\). This seems right. My application for 23 seems right too. The options might be wrong. Let me craft a question where the answer is 1. I need \((p-2)! \pmod p\). So let's ask for \(21! \pmod {23}\).