Prove that for every positive integer n there are n consecut

Prove that for every positive integer n, there are n consecutive composite integers. Example: for n = 3, we verify that 4! + 2 = 26, 4!+ 3 = 27, 4!+ 4 = 28 are composite numbers, where 4! = 4 middot 3 middot 2 middot 1.

Solution

Proof :-

Assume m and n exists in the positive integers and that m is less than n.

if m < n

n!=123mn

which is to say that m is a factor of n!.

So m+n! = m+(12mn) = m[(12n)/m+1]

m is a factor of n! and so n!/m is still an integer.

So since m is an integer that is bounded between 1 and n,

it stands that whatever number you pick up to n can divide m+n! making it composite till the n^th integer,

but n! has that n^th integer in it so the n^th integer is also composite which means that you can pick any integer between 1 and n inclusively and it will be composite.

Example :-

if n = 4

4! = 4.3.2.1 = 24

now pick any number from 1 to n so m = 1,2,3,4

so m+n! = 25,26,27,28

and these are divided by 1,2,3,4