what is the remainder of n! when divided by n+1 when n+1 is a prime number?

Answer 1

Try a few:

1! / 2 = 1 / 2 = 0 r 1
2! / 3 = 2 / 3 = 0 r 2
4! / 5 = 24 / 5 = 4 r 4
6! / 7 = 720 / 7 = 102 r 6
10! / 11 = 3628800 / 11 = 329890 r 10

I'd say the answer is n.

Answer 2

It's n. I think this is called "Wilson's Theorem." Here's a way to prove it. Say m is any integer less than n+1. Then consider the sequence of numbers m mod n+1, m^2 mod n+1,m^3 mod n+1, ... , m^k mod n+1, .... If you're not familiar with the "mod" notation, this just means the remainder when you divide by n+1. Anyway, since n+1 is prime and m is less than n+1, m^k mod n+1 is never 0, but always some integer from 1 to n (from here on I'll call these "the integers less than n+1" even though this isn't quite correct). Thus eventually there must be two powers of m, say m^k and m^l that are equal mod n+1. Say also that k>l. This means

m^k = m^l mod n+1
m^k - m^l = 0 mod n+1
m^l*(m^(k-l) - 1) = 0 mod n+1
m^(k-l) = 1 mod n+1 (since m^l doesn't = 0 mod n+1)

In particular this means that for every integer m less than n+1 there is another integer s (s = m^(k-l-1)) less than n +1 such that m*s = 1 mod n+1. We say that s is m inverse mod n+1.

Now consider:

x^2 = 1
x^2 - 1 = 0
(x+1)*(x-1) = 0

If you take these equations to be mod n+1, this shows that the only integers less than n +1 satisfying x^2 = 1 mod n+1 are 1 and n. Now you can prove the result as follows. Consider

n! = 1*(2* ... *(n-1))*n

Every number in the parentheses has an inverse mod n+1 somewhere in the parentheses. Thus when you pair these up you just get:

n! = 1*(1*1*...*1)*n = n mod n+1

Answer 3

Example:

let n= 4
Then n! = 24, and n+1 = 5 (which is prime)
So 24/4 has a remainder of n = 4
Try n=6. Then n! = 720 and n+1 = 7
So 720/7 = 102 with remainder n= 6.
The remainder will always be n.

Answer 4

It is called "Wilson's Theorem" that if p is prime, then when (p-1)! is divided by p, the remainder is p-1.

By substituting n+1 for p, the answer to your question is that the remainder is n, as other answerers have said - but they did not say WHY.

If the Mathworld Wolfram page is too difficult for you, just Google for "Wilson's Theorem" and you should find something simpler somewhere else.

Answer 5

if we divide n! by (n+1)!, where n+1 is a prime number, we use to get approximately n as the reminder.
for example
lets take 10 = n
then 11 is a prime number = n+1
if 10!/11 then we will get 329890 as the qotient & 10 as the reminder
where 10! = (329890*11)+10
if we take 12=n
then 13 is a prime number = n+1
if 12!/13 then we will get 36846276 as the qotient & 12 as the reminder
where 12! = (36846276*13)+13

Answer 6

Well, 5 is a prime number and the remainder of 5!/6 is 0.
And 7 is a prime number and the remainder of 7!/8 is 0.

Answer 7

n / (n + 1) = 0 remainder n

Answer: n is the remainder.

Answer 8

n/4

Answer 9

1

as n+1-(1)(n)=1

so remainder = 1
and qu't = 1

Answer 10

the answer is n.
dont know the proof.