Let n be any natural number greater than 1. Explain why the numbers n! 2, n! 3, n! 4, ..., n! n must all be composite. (This exercise shows that it is possible to find arbitrarily long sequences of consecutive composite numbers.)

Because each term of the sequence generates numbers with more than 1 and itself as dividers

Step-by-step explanation:

Just for the sake of correction.

1) Let's consider that

n! =n(n-1)(n-2)(n-3)...

2)And examine some numbers of that sequence above:

Every Natural number plugged in n, and added by two will a be an even number not only divisible by two, but in some cases by other numbers for example,n=4, then 4!+2=26 which has four dividers.

3) Similarly, the same happens to

and

Where we can find many dividers.

There's an example of a sequence, let's start with a prime number greater than 1

Let n=11

That's a long sequence of consecutive composite numbers, n=11.

substitute,

circle 1 = 5

circle 2 = 20

circle 3 = 44

in this case no matter how you add, you will always get a perfect square..

hence proved

happy to

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