Number Theory 6 Successive numbers no one is prime

[Amer]

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]

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]

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]

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]

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]

Thanks very much, you are amazing. (f)

 

[Amer]

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
[tex]n +2 = 2.3.4...k +2 = 2(3.4...k +1 ) [/tex] not prime
[tex]n+3 = 2.3.4...k + 3 = 3(2.4...k+1) [/tex] not prime
Thanks

 

[soroban]

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: .[tex]x \:=\:7!+2[/tex]

. . [tex]\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}[/tex]


There is a simpler (and much longer) list:

. . [tex]\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}[/tex]

 

[Amer]

Hello, Amer!


One solution is: .[tex]x \:=\:7!+2[/tex]

. . [tex]\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}[/tex]


There is a simpler (and much longer) list:

. . [tex]\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}[/tex]

Thanks, and Hello
we can make a list with k numbers which are not prime like that
[tex]k! + i [/tex] i=1,...,k