Exercise · ZW-WILSON-S-THEOREM-008 · ★★☆ Intermediate
What is \((p-1)! + 1\) modulo \(p\) for \(p = 5\)?
Choose one option.
Show answer
- C. \(0\)
Explanation
Since \(4! \equiv -1 \pmod{5}\), adding \(1\) gives a multiple of \(5\), so the remainder is \(0\).
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-006 If n>1 satisfies (n-1)!≡ -1 pmodn, what can you conclude about n?
- ZW-WILSON-S-THEOREM-007 Modulo 7, which elements of \1,2,dots,6\ are their own inverses?
- ZW-WILSON-S-THEOREM-009 Knowing that 12! ≡ -1 pmod13, find the remainder of 11! when divided by 13.