Exercise · ZW-WILSON-S-THEOREM-006 · ★★☆ Intermediate
If \(n>1\) satisfies \((n-1)!\equiv -1 \pmod{n}\), what can you conclude about \(n\)?
Choose one option.
Show answer
- A. \(n\) is prime
Explanation
The converse also holds: if \(n>1\) satisfies the congruence, \(n\) is prime, because a composite \(n\) has a proper divisor that appears among the factors of \((n-1)!\).
Review the concept: Wilson's 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
- ZW-WILSON-S-THEOREM-001 What does Wilson's theorem state?
- ZW-WILSON-S-THEOREM-002 What is the remainder of 4! when divided by 5?
- ZW-WILSON-S-THEOREM-003 In Wilson's theorem, what condition must p satisfy?
- ZW-WILSON-S-THEOREM-004 For p = 3, what is 2! modulo 3?
- ZW-WILSON-S-THEOREM-005 What does (p-1)! ≡ -1 pmodp mean?
- ZW-WILSON-S-THEOREM-007 Modulo 7, which elements of \1,2,dots,6\ are their own inverses?
- ZW-WILSON-S-THEOREM-008 What is (p-1)! + 1 modulo p for p = 5?
- ZW-WILSON-S-THEOREM-009 Knowing that 12! ≡ -1 pmod13, find the remainder of 11! when divided by 13.