-
Jan 17th, 2008, 10:53 AM
#1
Thread Starter
New Member
m!+ K
I Have been given this question and cannot get an answer so here goes if anyone can help within the next hour would be great.
Explore the divisors of m! + k for various m and k. Hence explain how to find a run of n consecutive integers which are composite.
Example:
m=5
5! = 120 Composite
5! + 1 = 121 Composite
5! + 2 = 122 Composite
5! + 3 = 123 Composite
5! + 4 = 124 Composite
5! + 5 = 125 Composite
5! + 6 = 126 Composite
5! + 7 = 127 Not Composite
The ? are suppose to be arrows.
-
Jan 17th, 2008, 11:17 PM
#2
Re: m!+ K
[The only truly interesting m are all greater than 1, so assume m>1]
When k=0, m!+k is of course composite.
In the example you had, you can also write 5!+3 as 5*4*3*2*1 + 3 = 3*(5*4*2*1 + 1) = 3*(5!/3 + 1) so 5!+3 is composite. In fact, you can write any of these m!+k as k*(m!/k+1). Here, 3 | 5!, so m!/k + 1 is an integer. So long as k > 1 and k | m!, you can factor m!+k like this into two numbers greater than 1, making m!+k composite.
You should probably be able to find the n consecutive integers using the ideas above. If not, tell me what you get and I'll fill in some more.
The time you enjoy wasting is not wasted time.
Bertrand Russell
<- Remember to rate posts you find helpful.
Posting Permissions
- You may not post new threads
- You may not post replies
- You may not post attachments
- You may not edit your posts
-
Forum Rules
|
Click Here to Expand Forum to Full Width
|