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, 04:50 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, 07:53 PMchiroRe: 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? - Oct 24th 2012, 10:42 PMSalahuddin559Re: prime number problem
Consider p = 5, you will get the answer.

Salahuddin

Maths online - Oct 25th 2012, 01:40 AMinoidRe: prime number problem
- Oct 25th 2012, 05:24 AMSalahuddin559Re: 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