Can Every Positive Integer Sequence Contain Only Composite Numbers?

  • Thread starter Thread starter NightFire
  • Start date Start date
NightFire
Messages
11
Reaction score
0
Hey everyone I am new in this forum, and it looks great ! :)
i have this problem to prove this question.id really appreciate our help for this one:

Prove that for every positive integer n,there are n consecutive composite integers.
Theres a the hint in the question that says; consider the n consecutive integers starting with (n+1)! + 2

thanks
for ur help
Roy
 
Physics news on Phys.org
Okay, (n+1)!+ 2 is composite because it is divisible by 2. (do you see why?)

(n+1)!+ 3 is composite because it is divisible by 3.

(n+1)!+ 4 is composite because it is divisible by 4.

etc.
 
HallsofIvy said:
Okay, (n+1)!+ 2 is composite because it is divisible by 2. (do you see why?)

(n+1)!+ 3 is composite because it is divisible by 3.

(n+1)!+ 4 is composite because it is divisible by 4.

etc.

i don't see why it is divisible by 2, would you like to explain it please?
much appreciated
 
Last edited:
Are you familiar with the factorial?
k!=k*(k-1)*(k-2)*...*3*2*1

Do you now see why it is divisible by 2?
 
This thread doesn't belong in this forum.

Is this a homework question? If so, this really should be in the "Homework/Coursework Questions" forum.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top