Question

show that 18! is congruent to -1 in modulo 437 ?

Answer 1

-1 mod 437 I believe equals 436; 436/2 = 218

Answer 2

idk how they are congruent

Answer 3

is there more to your question

Answer 4

it is solve through wilson's theorem which is (an integer p is a prime if and only if (p-1)! is congruent to -1 in modulo p )

Answer 5

I'm noticing that

23*19 = 437

but I'm not sure how to fit it in

Answer 6

do you have the original question to post

Answer 7

23 and 19 both divide 218!
Since they are primes, their product 23*19 also divides 218!

Answer 8

wouldn't it be like this @ganeshie8 ?

218! = 218*217*...*24*23*22*21*20*19*18...*3*2*1

218! = 218*217*...*24*(23*19)*22*21*20*18...*3*2*1

218! = 218*217*...*24*(437)*22*21*20*18...*3*2*1

218! = 437*218*217*...*24*22*21*20*18...*3*2*1

but since 437 is a factor, this means 218! = 0 (mod 437) so I'm not sure if I messed up somewhere or if the original statement `218! = -1 (mod 437)` is incorrect?

Answer 9

Exactly
The given congruence is false

Answer 10

Maybe they accidentally thought 437 was prime haha

Answer 11

good point @Kainui I thought it was too until I used a prime number checker

Answer 12

sorry guys its 18! not 218!

Answer 13

Here is one way to do it (see the attached text document). It's probably one of the longer ways possible. If not the longest route.

There is probably a much more clever way to do this. I can't think of it right now.

Answer 14

Wilson gives us
18! = -1 (mod 19)
22! = -1 (mod 23)

Answer 15

Notice that
22! = 18!*19*20*21*22
= 18!*(-4)(-3)(-2)(-1)
= 18!*24
= 18! (mod 23)

Answer 16

That means we have
18! = -1 (mod 19)
18! = -1 (mod 23)

Since 19 and 23 are primes, 19*23 also divides 18!+1