A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Just another cute number theory problem:
\((1)\) Find the remainder when \(100!^{100!}+99\) is divided by \(101\).
PS: This problem is posted on request of @Limitless
anonymous
 4 years ago
Just another cute number theory problem: \((1)\) Find the remainder when \(100!^{100!}+99\) is divided by \(101\). PS: This problem is posted on request of @Limitless

This Question is Closed

mathmagician
 4 years ago
Best ResponseYou've already chosen the best response.0first, you should see that 100^k mod 101 is equal to 100, if k is odd and 1 if k is even. The same happens with 100!^k. So, 100!^100!=1mod 101 and 100!^100!+99=100 mod 101

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0By Wilson's theorem, \[ 100! \equiv 1 \pmod {101} \] Since \(100!\) is even: \[100!^ {100!} \equiv 1 \pmod {101} \implies 100!^ {100!} +99 \equiv 100 \pmod {101} \] Hence, \(100\) is the remainder. Ref: http://en.wikipedia.org/wiki/Wilson's_theorem

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0EDIT: \[ 100!^ {100!} \equiv 1 \pmod {101} \]\[\implies 100!^ {100!} +99 \equiv 100 \pmod {101} \]

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.0You don't even need Wilson's Theorem for this. Since \(100!\) is a multiple of \(100\), and coprime to 101, we have that \(100!^{100!}\equiv1\pmod{101}\) by Fermat's Little Theorem. Hence, \(100!^{100!}+99\equiv100\pmod{101}\).
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.