If you know that 9001 is prime, what remainder does 8999! - 1 give when divided by 9001?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
if p is a prime no. then (p-2)! %p gives remander 1 ( wilson theorem)
Wilson's theorem states that for every prime p, p divides (p-2)! -1 (or equivalently (p-1)! + 1) .Since 9001 is prime, we can use the theorem and voila!
In fact, Wilson's theorem states that ( p − 1 ) ! ≡ − 1 ( m o d p ) , for any prime p .
Therefore, 9 0 0 0 ! ≡ − 1 m o d 9 0 0 1 ≡ 9 0 0 0 m o d 9 0 0 1 .
Dividing by 9000, we have 8 9 9 9 ! ≡ 1 m o d 9 0 0 1 ⇒ 8 9 9 9 ! − 1 ≡ 0 m o d 9 0 0 1
Log in to reply
Does division, as you said (dividing by 9000), always work?
Log in to reply
Yes it always works, because (p -1,p)=1 and thus we can divide the modular equation.In this case, (9000,9001)=1 as it should be, and dividing by a number coprime to 9001 is allowed in modular equations.
Log in to reply
@Bogdan Simeonov – That shows the equivalence of p dividing (p-2)! - 1 and (p-1)! + 1
It is equivalent to saying p divides (p-2)! - 1 .I wrote it down on my solution.I just used the (p-2)! one because it is more straightforward.
Problem Loading...
Note Loading...
Set Loading...
Usually Wilson's theorem states that ( p -1)! ≡ − 1 mod p , however we can manipulate this as follows: let x be the remainder when 8999! is divided by 9001, then 9000 x ≡ − x ≡ − 1 mod(9001). ∴ x =1 or in other words 8999! ≡ 1 mod (9001)