1. ## 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

2. Hello,

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

Originally Posted by kel1487
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
$\frac{31!}{22!}=31*30*29*28*27*26*25*24*23$

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

3. Originally Posted by Moo
$\frac{31!}{22!}=31*30*29*28*27*26*25*24*23$
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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