Prove that for every positive integer n, there are n consecutive compositie integers?

Question

A hint was given to consider the n consecutive integers starting with (n+1)! + 2... but I'm not entirely sure where to go with this problem...

Answer 1

Consider 6! = 1*2*3*4*5*6
It is divisible by 1, 2, 3, 4, 5 and 6

6! + 1 will be divisible by 1 (being divisible by 1 does NOT imply that the number will be composite)
6! + 2 will be divisible by 2
6! + 3 will be divisible by 3
6! + 4 will be divisible by 4
6! + 5 will be divisible by 5
6! + 6 will be divisible by 6

6!+2....6!+6 is a series if 5 consecutive composite integers. This can be generalised for any integer.

Answer 2

Going off the hint, consider those n consecutive integers:

(n+1)! + 2, (n+1)! + 3, (n+1)! + 4, ..., (n+1)! + n+1.

Convince yourself that these are divisible by 2, 3, 4, ..., n+1 respectively. Thus, we have found the desired n consecutive composite integers. This is sort of an intuitive explanation. To make it more rigorous you can use induction.

Answer 3

Composite Integer

Answer 4

Induction is NOT needed. Just assume that k divides (n+1)! for all k between 1 and n+1, as per the definition of (n+1)!. Then look at the n consecutive integers (n+1)! + 2, ... (n+1)! + (n+1).

Answer 5

Although you have not seen it, the idea that leads to the solution of this problem is quite simple and easy to comprehend. First try to factorize the given expression; n^3 - n = (n-1)(n)(n+1) Now let us exmine the 3 factors; (n-1), n and (n+1) are three consecutive intergers. Therefore 3 will divide evenly into any one of them. For instnce, take the three numbers 28, 29 and 30. As you see 3 will divide evenly into 30. Once again take 1001, 1002 and 1003 and you will see that 3 will divide evenly into 1002. If one of the factor of an expression can be devided by 3 with no remainder, 3 will divide evenly into the expression. If you want to know more about these things try to get hold of book on number theory. Have a nice day!