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Math Help - Wilson's Theorem applications

  1. #1
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    Wilson's Theorem applications

    Use Wilson's Theorem to find the least nonnegative residue modulo m of each integer n below
    a. 64!=n and m=67
    b. 31!/22!=n and m=11

    Also,

    Prove that if p is an odd prime number, then 2(p-3)! congruent to -1modp

    I missed class due to a wedding and our book only states wilson's theorem...so im really confused..thanks for any help1
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  2. #2
    Moo
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    Hello,

    Keep in mind Wilson's Theorem : (p-1)!=-1 \bmod p

    Quote Originally Posted by kel1487 View Post
    Use Wilson's Theorem to find the least nonnegative residue modulo m of each integer n below
    a. 64!=n and m=67
    64!=? \bmod 67

    We know by the theorem that 66!=-1 \bmod 67.

    But 66!=64! \times 65 \times 66. Hey ! 66=-1 \bmod 67 and 65=-2 \bmod 67

    So we have 64! \times 2=-1 \bmod 67

    --> \boxed{64!=-a \bmod 67}, where a is the inverse of 2 modulo 67, and such that 2a=1 \bmod 67

    One can see that 67=1+66=1+2*33 \implies 2*(-33)=1 \bmod 67

    \therefore a=-33 \bmod 67

    Therefore \boxed{64!=33 \bmod 67}


    b. 31!/22!=n and m=11
    \frac{31!}{22!}=31*30*29*28*27*26*25*24*23
    Let me think about it

    Also,

    Prove that if p is an odd prime number, then 2(p-3)! congruent to -1modp
    See the first part of a/

    (p-1)!=(p-3)!*(p-2)*(p-1)

    I missed class due to a wedding and our book only states wilson's theorem...so im really confused..thanks for any help1
    I hope the wedding was ok
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  3. #3
    Moo
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    Quote Originally Posted by Moo View Post
    \frac{31!}{22!}=31*30*29*28*27*26*25*24*23
    Let me think about it
    31=9 mod 11
    30=8 mod 11
    29=7 mod 11
    *etc*
    23=1 mod 11

    Thus 31!/22!=9! mod 11=10!/10 mod 11

    Work it out like previously
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