please help me

p is a prime.

given,

(p-1)! = p-1 (mod k)

k = ...

a. p+1

b. $\displaystyle \frac{(p-1)p}{2}$

c. $\displaystyle \frac{(p+1)p}{2}$

d. p+1

e. p^{2}

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- Oct 24th 2012, 05:22 PMinoidprime number problem
please help me

p is a prime.

given,

(p-1)! = p-1 (mod k)

k = ...

a. p+1

b. $\displaystyle \frac{(p-1)p}{2}$

c. $\displaystyle \frac{(p+1)p}{2}$

d. p+1

e. p^{2} - Oct 24th 2012, 08:02 PMjohnsomeoneRe: prime number problem
*IF* those are the only choices (I haven't checked that the only possible solution is correct), then just consider what happens when p=5.

Choices a and d are the same. Are you sure there there's no mistake in copying the problem? (This would seem to be asking you to use Wilson's Theorem.) - Oct 25th 2012, 01:31 AMinoidRe: prime number problem
sorry,

d. p-2

i don't know how to use wilson's theorem - Oct 25th 2012, 10:43 AMjohnsomeoneRe: prime number problem
Wilson's Theorem: (p-1)! = -1 (mod p) if and only if p is a prime.

1) It's what's "expected" to be used to answer this.

The use is that, if p is a prime, then -1 = (p-1)! = (p-1)(p-2)! = -(p-2)! (mod p), so that (p-2)! = 1 (mod p), so p divides [(p-2)!-1].

Ask yourself which k could make this fraction an integer: $\displaystyle \frac{(p-1)! - (p-1)}{k} = \frac{(p-1) \ [(p-2)! - 1]}{k}$

2) The other approach to this problem doesn't involve a real mathematical understanding, but rather understanding a trick for how to answer math questions on a test. So, this isn't the ideal way to do it, but it is a way, given that you trust that your teacher gave you a "correct" problem that didn't have an error.

When you're given a multiple choice question "for any prime p", then the answer has to be the same "for any prime p". In particular, the answer has to work for the prime p = 5. Only one of those choies, (a)-(e), holds when p=5, and so that choice must be the correct answer.