The discussion revolves around the possibility of finding six consecutive integers such that none of them is prime. Participants explore various examples and generalizations related to this problem, including specific values and methods for generating such sequences.
Participants do not reach a consensus on a single method or solution. Multiple competing views and examples are presented, indicating that the discussion remains unresolved regarding the best approach or the smallest solution.
The discussion includes various assumptions about the properties of numbers and the conditions under which they are considered non-prime. The effectiveness of the proposed methods may depend on specific definitions and contexts.
Amer said: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 said: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?...
chisigma said: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
$\chi$ $\sigma$
One solution is: .x \:=\:7!+2Is 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!
soroban said:Hello, Amer!
One solution is: .x \:=\:7!+2
. . \begin{array}{c}7!+2\text{ is divisible by 2} \\ <br /> 7!+3\text{ is divisible by 3} \\ 7!+4\text{ is divisible by 4} \\ <br /> 7!+5\text{ is divisible by 5} \\ 7!+6 \text{ is divisible by 6} \\ <br /> 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}