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FoolForMath

  • 3 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

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  1. .Sam.
    • 3 years ago
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    Wilson's Theorem?

  2. FoolForMath
    • 3 years ago
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    Bingo! ^^

  3. mathmagician
    • 3 years ago
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    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

  4. FoolForMath
    • 3 years ago
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    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

  5. FoolForMath
    • 3 years ago
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    EDIT: \[ 100!^ {100!} \equiv 1 \pmod {101} \]\[\implies 100!^ {100!} +99 \equiv 100 \pmod {101} \]

  6. KingGeorge
    • 3 years ago
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    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, \(100!^{100!}+99\equiv100\pmod{101}\).

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