# prime number problem

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• October 24th 2012, 05:50 PM
inoid
prime number problem
please help me
p is a prime.
given,
(p-1)! = p-1 (mod k)
k = ...
a. p+1
b. $\frac{(p-1)p}{2}$
c. $\frac{(p+1)p}{2}$
d. p+1
e. p2
• October 24th 2012, 08:53 PM
chiro
Re: prime number problem
Hey inoid.

If (p-1)! = (p-1) (mod k) then (p-1)! - (p-1) = 0 mod (k) which means (p-1)! - (p-1) | k or (p-1)*([p-2)! - 1] = kn for some integer n. Can you complete the rest?
• October 24th 2012, 11:42 PM
Salahuddin559
Re: prime number problem
Consider p = 5, you will get the answer.

Salahuddin
Maths online
• October 25th 2012, 02:40 AM
inoid
Re: prime number problem
Quote:

Originally Posted by chiro
Hey inoid.

If (p-1)! = (p-1) (mod k) then (p-1)! - (p-1) = 0 mod (k) which means (p-1)! - (p-1) | k or (p-1)*([p-2)! - 1] = kn for some integer n. Can you complete the rest?

i can't....
please.......
• October 25th 2012, 06:24 AM
Salahuddin559
Re: prime number problem
Ok, don't worry, I will explain.

If p = 5, then (First of all convince yourself that 5 is a prime number. If you cant understand that, dont worry, please ask, I will explain that also).
p - 1 = 4, (p - 1)! = 4! = 1 * 2 * 3 * 4 = 2 * 3 * 4 = 6 * 4 = 24.

It says to find k, when (24 = 4) mod k.

The given options are:

(p + 1) = (5 + 1) = 6.
(p - 1) * p / 2 = (5 - 1) * 5 / 2 = 4 * 5 / 2 = 20 / 2 = 10
(p + 1) * p / 2 = (5 + 1) * 5 / 2 = 6 * 5 / 2 = 30 / 2 = 15.
p^2 = 5^2 = 25.

Take modulo for each of these numbers and check if 24 and 4 give the same moduli, with these numbers. Whichever number (among 6, 10, 15 and 25) gives same modulus, that is your answer k.

Salahuddin
Maths online