m!+ K

[Delta88]

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.

[jemidiah]

[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.