Exercise · ZW-WILSON-S-THEOREM-010 · ★★★ Advanced
For \(n = 4\), does \((n-1)! \equiv -1 \pmod{n}\) hold? Compute \(3!\) modulo \(4\) and compare with \(-1\) modulo \(4\).
Choose one option.
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- A. No, because \(3! \equiv 2\) and \(-1 \equiv 3 \pmod{4}\)
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Explanation
\(3! = 6 \equiv 2 \pmod{4}\), and \(-1 \equiv 3 \pmod{4}\), so the congruence fails and \(4\), being composite, does not satisfy the theorem.
For a prime number \(p\), \((p-1)!\) leaves remainder \(-1\) when divided by \(p\).
\[(p-1)!\equiv -1 \pmod{p}\]
More practice problems: wilson's theorem