Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
Wilson's Theorem?
Bingo! ^^
first, 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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

By 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
EDIT: \[ 100!^ {100!} \equiv 1 \pmod {101} \]\[\implies 100!^ {100!} +99 \equiv 100 \pmod {101} \]
You 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, \(100!^{100!}+99\equiv100\pmod{101}\).

Not the answer you are looking for?

Search for more explanations.

Ask your own question