Exercise · ZW-WILSON-S-THEOREM-002 · ★★☆ Intermediate
What is the remainder of \(4!\) when divided by \(5\)?
Choose one option.
Show answer
- A. \(4\)
Explanation
\(4! = 24 = 5\cdot 4 + 4\), so the remainder is \(4\), which is equivalent to \(-1\) modulo \(5\).
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-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-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.