Discussion

What is the remainder when \((20!)\) is divided by 23?

Answer: A

Correct answer: 11

Wilson’s theorem states that \((p-1)! \equiv -1 \pmod p\) for any prime \(p\). With \(p = 23\),

\[22! \equiv -1 \pmod{23}\]

Peel off the two largest factors, using \(22 \equiv -1\) and \(21 \equiv -2 \pmod{23}\):

\[22! = 22\times 21\times 20! \equiv (-1)(-2)\,20! = 2\cdot 20! \pmod{23}\]

So \(2\cdot 20! \equiv -1 \equiv 22 \pmod{23}\), and halving gives \(20! \equiv 11 \pmod{23}\).

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