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True/False

If $p$ is a $3$-digit prime, then there always exist $p$ consecutive composite numbers.

How to approach this?

  • 2
    A more difficult question is whether there exist $p$ consecutive composite numbers preceded and followed by primes. Without the "3 digit" restriction this is not known. With it, it is a matter for explicit computation (see http://www.trnicely.net/gaps/gaplist.html#MainTable)2012-07-08

3 Answers 3

4

$P_k =(s + 1)! + k$ for $k = 2$ to $(s + 1)$

are $s$ consecutive positive integers

as $P_k$ is always divisible by $k$, hence the statement is true

4

Hint $\rm\ 1 are composite. Now put $\rm\,a,b,c = 2,3,4\ldots$ and $\rm n =\,$ __

-1

The statement is false!
The first 3 digit prime is 101, the next prime is 103; difference is 2 not 101! Take another 3 digit prime, 131 the next prime is 137; difference is 6 not 131! You can look up any of the 3 digit primes and by inspection the difference is well under 100. Ron

  • 4
    It seems to me that you have grossly misread the questio$n$. -12012-07-08