Wilson's Theorem applications

• Jul 21st 2008, 01:29 PM
kel1487
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
• Jul 21st 2008, 01:47 PM
Moo
Hello,

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

Quote:

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

$\displaystyle 64!=? \bmod 67$

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

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

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

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

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

$\displaystyle \therefore a=-33 \bmod 67$

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

Quote:

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

Quote:

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)

Quote:

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 (Rofl)
• Jul 23rd 2008, 07:22 AM
Moo
Quote:

Originally Posted by Moo
$\displaystyle \frac{31!}{22!}=31*30*29*28*27*26*25*24*23$
Let me think about it :p

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 :)