For any given natural number 'n', prove that there always exist 'n' consequtive natural numbers that are composite.INFO: Any number greater than 1 which is not a prime is known as a composite number.

embizze | Certified Educator

Given `n in NN,n>1` (n an integer greater than 1), prove that there is a string of n composite numbers.

For a given n, use the following sequence of numbers:

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

First note that there are n numbers in the sequence.

Every number in this sequence is composite. The first number, `(n+1)!+2`, is divisible by 2. (if a is divisible by , and b is divisible by 2, then a+b is divisible by 2 -- by definition (n+1)! is divisible by every number smaller than (n+1)!)

The second number is a multiple of 3, etc... The last number is a multiple of n+1, thus you have a sequence of n numbers that are all composite.

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Example: let n=5. The sequence generated is 6!+2,6!+3,6!+4,6!+5,6!+6. Now 6!=6*5*4*3*2*1=720 so the sequence is 722,723,724,725,726. 722 is a multiple of 2, 723 is a multiple of 3, 724 is a multiple of 4, 725 is a multiple of 5, and 726 is a multiple of 6. This works for any n.