Prove that none of them is prime

[kmeado07]

Homework Statement

Let n be a part of the natural numbers, with n>=2. Consider the numbers [n factorial +2 ], [n factorial + 3], ..., [n factorial + n].
Prove that none of them is prime, and deduce that there are arbitrarily long finite stretches of consecutive non-prime nummbers in the natural numbers.

Homework Equations

The Attempt at a Solution

I really don't know how to do this question. Any help/hints would be appreciated.

 

[tiny-tim]

Let n be a part of the natural numbers, with n>=2. Consider the numbers [n factorial +2 ], [n factorial + 3], ..., [n factorial + n].
Prove that none of them is prime, and deduce that there are arbitrarily long finite stretches of consecutive non-prime nummbers in the natural numbers.

Hi kmeado07!

Hint: what factor is there of (n! + 3)?

 

[kmeado07]

Is (n factorial + 2) a factor of it?

Which would then apply to the others, which would show that none of them were prime.
For them each to be prime, their only factors would have to be themselves and 1.

 

[tiny-tim]

Hi kmeado07!

(please use the !)

Is (n factorial + 2) a factor of it?

No: n! + 2 is only one less than n! + 3 …

how can it be a factor of it?

Try writing n! + 3 in full (with n = 7, say)

 

[kmeado07]

ok, so a factor of n!+3 is 3, and a factor of n!+2 is 2 and so on.
So a factor of n!+n is n. This shows that none of them is prime.

How would i go about doing the second part of the question?

 

[tiny-tim]

Prove that none of them is prime, and deduce that there are arbitrarily long finite stretches of consecutive non-prime nummbers in the natural numbers.

ok, so a factor of n!+3 is 3, and a factor of n!+2 is 2 and so on.
So a factor of n!+n is n. This shows that none of them is prime.

How would i go about doing the second part of the question?

ok, so how many consecutive non-prime numbers are there starting with n!?

 

[kmeado07]

So there are n-2 consecutive non-primes starting with n! ?

 

[tiny-tim]

So there are n-2 consecutive non-primes starting with n! ?

ooh, i got that slightly wrong, didn't i?

yes, n-1 starting with n! + 2 …

so if you wanted 1,000,000 consecutive non-primes, where would you start?

 

[kmeado07]

you would start at n! + 1,000,003 ?

 

[tiny-tim]

you would start at n! + 1,000,003 ?

Nowhere near …

try again ​

 

[kmeado07]

i don't know, I am confused!

 

[Firepanda]

bump i'd like more info on this too

 

[HallsofIvy]

So there are n-2 consecutive non-primes starting with n! ?

so if you wanted 1,000,000 consecutive non-primes, where would you start?

you would start at n! + 1,000,003 ?

If n- 2= 1,000,000, what is n? It's that easy.

 

[Firepanda]

ooh, i got that slightly wrong, didn't i?

yes, n-1 starting with n! + 2 …

so if you wanted 1,000,000 consecutive non-primes, where would you start?

I really have no idea what's going on starting with this

are you saying for n=10, then

2 + 10*9*8*...*2*1

has n-1 = 9 consectuve non primes?

because I don't see that, maybe I am not understanding the question..

 

[D H]

Which numbers from 2 to 10 fails to divide 10! ? Given that, what divides 10! + 2? 10! + 3, and so on?