# Number Theory6 Successive numbers no one is prime

#### Amer

##### Active member
is it possible to find a 6 Successive numbers like
x , x+1 , x+2 , x+3 ,x+4 ,x+5 such that one one is prime ?

Thanks

#### chisigma

##### Well-known member
is it possible to find a 6 Successive numbers like
x , x+1 , x+2 , x+3 ,x+4 ,x+5 such that one one is prime ?

Thanks
It is comfortable to verify that $x=5!=120$ satisfies the request... and the reason of that is easy to see...

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
Of course $x=5!$ is not the only and neither the 'smallest' solution. Setting $x=90$ You have 7 consecutive non prime numbers...

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
Why don't try to generalize the problem: given k, how to compute an n such that n, n+1, n+2,...,n+k are all non prime numbers?...

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
Why don't try to generalize the problem: given k, how to compute an n such that n, n+1, n+2,...,n+k are all non prime numbers?...
An easy way to get the result in the particular case where k is prime is based on the consideration that $k|n \implies k|(n+m k)$. Setting $n = 2 \cdot 3 \cdot 5 \cdot ... \cdot k$ we are sure that $n+2,n+3,...,n+k+1$ are all non prime numbers. For example...

$\displaystyle k=11 \implies n=2310 \implies 2312,2313,2314,2315,2316,2317,2318,2319,2320,2321,2322\ \text{are all non prime }$

Although 'easy' this method is often 'excessive' because the effective quantity consecutive non prime numbers can be greater. In the given example 2311 is prime so that the sequence starts at 2312 but 2323,2324,2325,2326,2327,2328,2329,2330,2331 and 2332 are non prime numbers [2333 is prime...] and the effective sequence's length is 20 [not 11]...

Kind regards

$\chi$ $\sigma$

#### Amer

##### Active member
Thanks very much, you are amazing. (f)

#### Amer

##### Active member
An easy way to get the result in the particular case where k is prime is based on the consideration that $k|n \implies k|(n+m k)$. Setting $n = 2 \cdot 3 \cdot 5 \cdot ... \cdot k$ we are sure that $n+2,n+3,...,n+k+1$ are all non prime numbers. For example...

$\displaystyle k=11 \implies n=2310 \implies 2312,2313,2314,2315,2316,2317,2318,2319,2320,2321,2322\ \text{are all non prime }$

Although 'easy' this method is often 'excessive' because the effective quantity consecutive non prime numbers can be greater. In the given example 2311 is prime so that the sequence starts at 2312 but 2323,2324,2325,2326,2327,2328,2329,2330,2331 and 2332 are non prime numbers [2333 is prime...] and the effective sequence's length is 20 [not 11]...

Kind regards

$\chi$ $\sigma$
nice one I get it
$$n +2 = 2.3.4...k +2 = 2(3.4...k +1 )$$ not prime
$$n+3 = 2.3.4...k + 3 = 3(2.4...k+1)$$ not prime
Thanks

#### soroban

##### Well-known member
Hello, Amer!

Is it possible to find a 6 consecutive integers numbers like
x , x+1 , x+2 , x+3 ,x+4 ,x+5 such that not one is prime? . Yes!

One solution is: .$$x \:=\:7!+2$$

. . $$\begin{array}{c}7!+2\text{ is divisible by 2} \\ 7!+3\text{ is divisible by 3} \\ 7!+4\text{ is divisible by 4} \\ 7!+5\text{ is divisible by 5} \\ 7!+6 \text{ is divisible by 6} \\ 7!+7\text{ is divisible by 7} \\ \end{array}$$

There is a simpler (and much longer) list:

. . $$\begin{array}{ccc} 114 &=& 2\cdot57 \\ 115 &=& 5\cdot 23 \\ 116 &=& 2\cdot 58 \\ 117 &=& 3\cdot39 \\ 118 &=& 2\cdot59 \\ 119 &=& 7\cdot17 \\ 120 &=& 2\cdot60 \\ 121 &=& 11\cdot11 \\ 122 &=& 2\cdot61 \\ 123 &=& 3\cdot41 \\ 124 &=& 2\cdot62 \\ 125 &=& 5\cdot25 \\ 126 &=& 2\cdot63 \end{array}$$

#### Amer

##### Active member
Hello, Amer!

One solution is: .$$x \:=\:7!+2$$

. . $$\begin{array}{c}7!+2\text{ is divisible by 2} \\ 7!+3\text{ is divisible by 3} \\ 7!+4\text{ is divisible by 4} \\ 7!+5\text{ is divisible by 5} \\ 7!+6 \text{ is divisible by 6} \\ 7!+7\text{ is divisible by 7} \\ \end{array}$$

There is a simpler (and much longer) list:

. . $$\begin{array}{ccc} 114 &=& 2\cdot57 \\ 115 &=& 5\cdot 23 \\ 116 &=& 2\cdot 58 \\ 117 &=& 3\cdot39 \\ 118 &=& 2\cdot59 \\ 119 &=& 7\cdot17 \\ 120 &=& 2\cdot60 \\ 121 &=& 11\cdot11 \\ 122 &=& 2\cdot61 \\ 123 &=& 3\cdot41 \\ 124 &=& 2\cdot62 \\ 125 &=& 5\cdot25 \\ 126 &=& 2\cdot63 \end{array}$$
Thanks, and Hello we can make a list with k numbers which are not prime like that
$$k! + i$$ i=1,...,k