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- Thread starter Amer
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- Feb 13, 2012

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

Kind regards

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

Kind regards

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

Kind regards

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

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...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?...

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

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

$\chi$ $\sigma$

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nice one I get itAn 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

$\chi$ $\sigma$

[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

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]

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Thanks, and HelloHello, 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]

we can make a list with k numbers which are not prime like that

[tex]k! + i [/tex] i=1,...,k