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does a number that contains all primes less than it exist?


Do there exist any dormant primes?Currently, what is the largest publicly known prime number such that all prime numbers less than it are known?Difference between two (not consecutive) primesProof that there exists a larger prime than prime number P, which is the largest of a finite set of primes?A very basic (I think) question about proofs and infinitely many primes.Can it be proven/disproven that there are highly composite numbers that prime-factorize into larger primes such as $9999991$?Given a number N, how many numbers less than N can be written as a product of less than k prime factors?Sums of consecutive primesIs it true for any $n=2p$ where $p$ is prime, that the number of twin primes less than $n$ approaches the number of prime pairs?Can it be shown that Double Goldbug Numbers cannot exist?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








6












$begingroup$


I want a number that has a prime factorization that contains all prime numbers less than that number (besides $2$), anyone with an answer please show a proof.



I have made a little progress, if this number exists, then it is one less than a prime number.

proof:

make a list of all the primes less than $n$ $P_1,space P_2,space P_3,cdots$
$n+1$ is not a factor of any of these primes so it ether is a prime or has a prime that our list missed and since the list contains all primes less than $n$, $n+1$ must be prime.



and more evidence that suggests this exists is the proof that any arbitrary prime gap exists. if $n!+1$ is prime then the prime gap between $n!+1$ and the next prime is at least $n-1$ because:
$n!+2$ is a multiple of $2$ and
$n!+3$ is a multiple of $3$ and so-on until
$n!+n$ which is a multiple of $n$



any more help would be appreciated










share|cite|improve this question









$endgroup$







  • 5




    $begingroup$
    Bertrand's postulate?
    $endgroup$
    – Lord Shark the Unknown
    8 hours ago










  • $begingroup$
    @LordSharktheUnknown if I understand Bertrand's postulate, no. he has a way of finding prime gaps (sort-of) I'm just wondering if a number can contain all prime numbers less that its self
    $endgroup$
    – spydragon
    8 hours ago











  • $begingroup$
    Prime gaps of arbitrary length appear. Small prime gaps persist indefinitely with gaps at most $246$ appearing infinitely often (en.wikipedia.org/wiki/Polymath_Project) - the conjecture is that $246$ can be reduced to $2$, and that "prime pairs" occur infinitely often. Primes are too dense for your idea to work. Exploring them is mathematically rich territory - proving new things about primes is almost always standing on the shoulders of giants.
    $endgroup$
    – Mark Bennet
    8 hours ago










  • $begingroup$
    You need to understand the question you are asking. Bertrand's postulate shows that there is a prime which is not "included" (because there is a prime between $n$ and $2n$ which is, therefore, not a factor of $2n$, and the product you are creating is of the from $2n$)
    $endgroup$
    – Mark Bennet
    8 hours ago










  • $begingroup$
    @MarkBennet I know, but as it hasn't been proven you and I cannot know for certain and even though primes are not always 1,000,000 numbers apart there could still be a number that works
    $endgroup$
    – spydragon
    8 hours ago

















6












$begingroup$


I want a number that has a prime factorization that contains all prime numbers less than that number (besides $2$), anyone with an answer please show a proof.



I have made a little progress, if this number exists, then it is one less than a prime number.

proof:

make a list of all the primes less than $n$ $P_1,space P_2,space P_3,cdots$
$n+1$ is not a factor of any of these primes so it ether is a prime or has a prime that our list missed and since the list contains all primes less than $n$, $n+1$ must be prime.



and more evidence that suggests this exists is the proof that any arbitrary prime gap exists. if $n!+1$ is prime then the prime gap between $n!+1$ and the next prime is at least $n-1$ because:
$n!+2$ is a multiple of $2$ and
$n!+3$ is a multiple of $3$ and so-on until
$n!+n$ which is a multiple of $n$



any more help would be appreciated










share|cite|improve this question









$endgroup$







  • 5




    $begingroup$
    Bertrand's postulate?
    $endgroup$
    – Lord Shark the Unknown
    8 hours ago










  • $begingroup$
    @LordSharktheUnknown if I understand Bertrand's postulate, no. he has a way of finding prime gaps (sort-of) I'm just wondering if a number can contain all prime numbers less that its self
    $endgroup$
    – spydragon
    8 hours ago











  • $begingroup$
    Prime gaps of arbitrary length appear. Small prime gaps persist indefinitely with gaps at most $246$ appearing infinitely often (en.wikipedia.org/wiki/Polymath_Project) - the conjecture is that $246$ can be reduced to $2$, and that "prime pairs" occur infinitely often. Primes are too dense for your idea to work. Exploring them is mathematically rich territory - proving new things about primes is almost always standing on the shoulders of giants.
    $endgroup$
    – Mark Bennet
    8 hours ago










  • $begingroup$
    You need to understand the question you are asking. Bertrand's postulate shows that there is a prime which is not "included" (because there is a prime between $n$ and $2n$ which is, therefore, not a factor of $2n$, and the product you are creating is of the from $2n$)
    $endgroup$
    – Mark Bennet
    8 hours ago










  • $begingroup$
    @MarkBennet I know, but as it hasn't been proven you and I cannot know for certain and even though primes are not always 1,000,000 numbers apart there could still be a number that works
    $endgroup$
    – spydragon
    8 hours ago













6












6








6





$begingroup$


I want a number that has a prime factorization that contains all prime numbers less than that number (besides $2$), anyone with an answer please show a proof.



I have made a little progress, if this number exists, then it is one less than a prime number.

proof:

make a list of all the primes less than $n$ $P_1,space P_2,space P_3,cdots$
$n+1$ is not a factor of any of these primes so it ether is a prime or has a prime that our list missed and since the list contains all primes less than $n$, $n+1$ must be prime.



and more evidence that suggests this exists is the proof that any arbitrary prime gap exists. if $n!+1$ is prime then the prime gap between $n!+1$ and the next prime is at least $n-1$ because:
$n!+2$ is a multiple of $2$ and
$n!+3$ is a multiple of $3$ and so-on until
$n!+n$ which is a multiple of $n$



any more help would be appreciated










share|cite|improve this question









$endgroup$




I want a number that has a prime factorization that contains all prime numbers less than that number (besides $2$), anyone with an answer please show a proof.



I have made a little progress, if this number exists, then it is one less than a prime number.

proof:

make a list of all the primes less than $n$ $P_1,space P_2,space P_3,cdots$
$n+1$ is not a factor of any of these primes so it ether is a prime or has a prime that our list missed and since the list contains all primes less than $n$, $n+1$ must be prime.



and more evidence that suggests this exists is the proof that any arbitrary prime gap exists. if $n!+1$ is prime then the prime gap between $n!+1$ and the next prime is at least $n-1$ because:
$n!+2$ is a multiple of $2$ and
$n!+3$ is a multiple of $3$ and so-on until
$n!+n$ which is a multiple of $n$



any more help would be appreciated







prime-numbers prime-factorization prime-gaps






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 8 hours ago









spydragonspydragon

4310 bronze badges




4310 bronze badges







  • 5




    $begingroup$
    Bertrand's postulate?
    $endgroup$
    – Lord Shark the Unknown
    8 hours ago










  • $begingroup$
    @LordSharktheUnknown if I understand Bertrand's postulate, no. he has a way of finding prime gaps (sort-of) I'm just wondering if a number can contain all prime numbers less that its self
    $endgroup$
    – spydragon
    8 hours ago











  • $begingroup$
    Prime gaps of arbitrary length appear. Small prime gaps persist indefinitely with gaps at most $246$ appearing infinitely often (en.wikipedia.org/wiki/Polymath_Project) - the conjecture is that $246$ can be reduced to $2$, and that "prime pairs" occur infinitely often. Primes are too dense for your idea to work. Exploring them is mathematically rich territory - proving new things about primes is almost always standing on the shoulders of giants.
    $endgroup$
    – Mark Bennet
    8 hours ago










  • $begingroup$
    You need to understand the question you are asking. Bertrand's postulate shows that there is a prime which is not "included" (because there is a prime between $n$ and $2n$ which is, therefore, not a factor of $2n$, and the product you are creating is of the from $2n$)
    $endgroup$
    – Mark Bennet
    8 hours ago










  • $begingroup$
    @MarkBennet I know, but as it hasn't been proven you and I cannot know for certain and even though primes are not always 1,000,000 numbers apart there could still be a number that works
    $endgroup$
    – spydragon
    8 hours ago












  • 5




    $begingroup$
    Bertrand's postulate?
    $endgroup$
    – Lord Shark the Unknown
    8 hours ago










  • $begingroup$
    @LordSharktheUnknown if I understand Bertrand's postulate, no. he has a way of finding prime gaps (sort-of) I'm just wondering if a number can contain all prime numbers less that its self
    $endgroup$
    – spydragon
    8 hours ago











  • $begingroup$
    Prime gaps of arbitrary length appear. Small prime gaps persist indefinitely with gaps at most $246$ appearing infinitely often (en.wikipedia.org/wiki/Polymath_Project) - the conjecture is that $246$ can be reduced to $2$, and that "prime pairs" occur infinitely often. Primes are too dense for your idea to work. Exploring them is mathematically rich territory - proving new things about primes is almost always standing on the shoulders of giants.
    $endgroup$
    – Mark Bennet
    8 hours ago










  • $begingroup$
    You need to understand the question you are asking. Bertrand's postulate shows that there is a prime which is not "included" (because there is a prime between $n$ and $2n$ which is, therefore, not a factor of $2n$, and the product you are creating is of the from $2n$)
    $endgroup$
    – Mark Bennet
    8 hours ago










  • $begingroup$
    @MarkBennet I know, but as it hasn't been proven you and I cannot know for certain and even though primes are not always 1,000,000 numbers apart there could still be a number that works
    $endgroup$
    – spydragon
    8 hours ago







5




5




$begingroup$
Bertrand's postulate?
$endgroup$
– Lord Shark the Unknown
8 hours ago




$begingroup$
Bertrand's postulate?
$endgroup$
– Lord Shark the Unknown
8 hours ago












$begingroup$
@LordSharktheUnknown if I understand Bertrand's postulate, no. he has a way of finding prime gaps (sort-of) I'm just wondering if a number can contain all prime numbers less that its self
$endgroup$
– spydragon
8 hours ago





$begingroup$
@LordSharktheUnknown if I understand Bertrand's postulate, no. he has a way of finding prime gaps (sort-of) I'm just wondering if a number can contain all prime numbers less that its self
$endgroup$
– spydragon
8 hours ago













$begingroup$
Prime gaps of arbitrary length appear. Small prime gaps persist indefinitely with gaps at most $246$ appearing infinitely often (en.wikipedia.org/wiki/Polymath_Project) - the conjecture is that $246$ can be reduced to $2$, and that "prime pairs" occur infinitely often. Primes are too dense for your idea to work. Exploring them is mathematically rich territory - proving new things about primes is almost always standing on the shoulders of giants.
$endgroup$
– Mark Bennet
8 hours ago




$begingroup$
Prime gaps of arbitrary length appear. Small prime gaps persist indefinitely with gaps at most $246$ appearing infinitely often (en.wikipedia.org/wiki/Polymath_Project) - the conjecture is that $246$ can be reduced to $2$, and that "prime pairs" occur infinitely often. Primes are too dense for your idea to work. Exploring them is mathematically rich territory - proving new things about primes is almost always standing on the shoulders of giants.
$endgroup$
– Mark Bennet
8 hours ago












$begingroup$
You need to understand the question you are asking. Bertrand's postulate shows that there is a prime which is not "included" (because there is a prime between $n$ and $2n$ which is, therefore, not a factor of $2n$, and the product you are creating is of the from $2n$)
$endgroup$
– Mark Bennet
8 hours ago




$begingroup$
You need to understand the question you are asking. Bertrand's postulate shows that there is a prime which is not "included" (because there is a prime between $n$ and $2n$ which is, therefore, not a factor of $2n$, and the product you are creating is of the from $2n$)
$endgroup$
– Mark Bennet
8 hours ago












$begingroup$
@MarkBennet I know, but as it hasn't been proven you and I cannot know for certain and even though primes are not always 1,000,000 numbers apart there could still be a number that works
$endgroup$
– spydragon
8 hours ago




$begingroup$
@MarkBennet I know, but as it hasn't been proven you and I cannot know for certain and even though primes are not always 1,000,000 numbers apart there could still be a number that works
$endgroup$
– spydragon
8 hours ago










5 Answers
5






active

oldest

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10












$begingroup$

This is impossible. For every $n>1$, we have gcd$(n,n-1)=1$. Since there is a prime dividing $n-1$, the result follows.



Edit : I should have assumed $n>2$ since I need a prime divisor of $n-1$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Well, $n=2$ actually works ...
    $endgroup$
    – Henning Makholm
    8 hours ago










  • $begingroup$
    @HenningMakholm you are correct, in $n=2$ case, $n-1$ is not divisible by any prime. I should have assumed that $n>2$.
    $endgroup$
    – Levent
    8 hours ago










  • $begingroup$
    what if $n-1$ and $n+1$ are twin primes, I never said $n$ had to be prime?
    $endgroup$
    – spydragon
    8 hours ago











  • $begingroup$
    @spydragon I did not assume that $n$ is a prime. Let $n>2$ be any integer. Then gcd$(n,n-1)=1$ (why?). Now $n-1>1$ so there exists a prime divisor $p$ of $n-1$ (it can be $n-1$ itself). Then $p<n$ and $p$ does not divide $n$ (since $p$ divides $n-1$ and gcd$(n,n-1)=1$).
    $endgroup$
    – Levent
    8 hours ago










  • $begingroup$
    @spydragon you want $n$ to contain every prime factor less than or equal to $n$ - it doesn't have to be prime. But the factors of $n-1$ don't divide $n$. If $n-1$ has a non-trivial prime factor, $n$ cannot contain it, Please think about this - try out some examples. $n=2$ is the only solution.
    $endgroup$
    – Mark Bennet
    8 hours ago


















6












$begingroup$

To make Lord Shark's comment explicit:



Let $n$ be composite and let $p$ be its largest prime factor. Then certainly $nge 2p$. By Bertrand's postulate, there exists a prime $q$ between $p$ and $2p$, hence there exist primes below $n$ that do not divide $n$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    unless that prime $q$ is in between $n$ and $2p$
    $endgroup$
    – spydragon
    8 hours ago










  • $begingroup$
    @spydragon $p$ was chosen as the largest prime divisor of $n$ so any larger prime - e.g. the $q$ that comes from Bertrand, or any other prime larger than $p$, does not divide $n$.
    $endgroup$
    – Ethan Bolker
    3 hours ago


















1












$begingroup$

$n=2cdotnover 2$ at most for the factoring to work out. But, by that very logic, applying it with Bertrand's Postulate $$exists cinmathbbP,nover 2leq cleq nimplies cmid n land 2nmid n$$ contradicting the possibility of an n with the property of divisibility by 2 working. which then says n doesn't include all primes less than it.






share|cite|improve this answer











$endgroup$




















    1












    $begingroup$

    Look at it from the other side, the "half-primorials" (see http://oeis.org/A070826) of which the first composite one is 15, which as you know is $3 times 5$.



    But it's not divisible by 7, 11 or 13. Multiply those in and you get 15015. But that's not divisible by 17, 19, 23, 29, ..., 14983 or 15013. Multiply those in... actually, don't, you might tie up your computer for too long.






    share|cite|improve this answer









    $endgroup$




















      0












      $begingroup$

      A small dilation on the answer of Levent: Let us imagine there is such a number and there are $n$ primes smaller than that number. Then the smallest number that contains all of those primes is the product of all of those primes, which is called a primorial and is denoted $p_n#$.



      Your question in effect asks, "Is it ever the case that $p_n+1>p_n#$?"



      We can start by considering the number $p_n#+1$. Since every one of the first $n$ primes divides $p_n#$, and none of them divide $1$, we conclude none of them can divide $p_n#+1$. So that number must either be a prime or have prime factors. If it has prime factors, they must be other than any of the first $p_n$, and those factors must be smaller than $p_n#$, and larger than $p_n$ itself. In this case, the prime next larger than $p_n$ would have to be smaller than $p_n#$ and the answer to your question would be "No". So that forces us to consider that $p_n#+1$ is in fact a prime, and postulate that it is the prime next larger than $p_n$, i.e. $p_n+1$. If it were ever the case that $p_n+1=p_n#+1$, that would be the only possible way to answer your question in the positive.



      Here is how Levent's answer sinks that ship. We must now consider the number $p_n#-1$. As in the case of $p_n#+1$, each $p_n$ divides $p_n#$ but none of them divide $1$, so $p_n#-1$ must either be a prime or have prime factors, in either instance smaller than $p_n#$. That being the case, $p_n#+1$ cannot be the prime next larger than $p_n$.



      Levent's answer stands on its own in answering your question in the negative. However, the excursion taken here shows that Levent's answer also undercuts the only reasonable candidate for the kind of number you postulate.






      share|cite|improve this answer









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        5 Answers
        5






        active

        oldest

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        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        10












        $begingroup$

        This is impossible. For every $n>1$, we have gcd$(n,n-1)=1$. Since there is a prime dividing $n-1$, the result follows.



        Edit : I should have assumed $n>2$ since I need a prime divisor of $n-1$.






        share|cite|improve this answer











        $endgroup$












        • $begingroup$
          Well, $n=2$ actually works ...
          $endgroup$
          – Henning Makholm
          8 hours ago










        • $begingroup$
          @HenningMakholm you are correct, in $n=2$ case, $n-1$ is not divisible by any prime. I should have assumed that $n>2$.
          $endgroup$
          – Levent
          8 hours ago










        • $begingroup$
          what if $n-1$ and $n+1$ are twin primes, I never said $n$ had to be prime?
          $endgroup$
          – spydragon
          8 hours ago











        • $begingroup$
          @spydragon I did not assume that $n$ is a prime. Let $n>2$ be any integer. Then gcd$(n,n-1)=1$ (why?). Now $n-1>1$ so there exists a prime divisor $p$ of $n-1$ (it can be $n-1$ itself). Then $p<n$ and $p$ does not divide $n$ (since $p$ divides $n-1$ and gcd$(n,n-1)=1$).
          $endgroup$
          – Levent
          8 hours ago










        • $begingroup$
          @spydragon you want $n$ to contain every prime factor less than or equal to $n$ - it doesn't have to be prime. But the factors of $n-1$ don't divide $n$. If $n-1$ has a non-trivial prime factor, $n$ cannot contain it, Please think about this - try out some examples. $n=2$ is the only solution.
          $endgroup$
          – Mark Bennet
          8 hours ago















        10












        $begingroup$

        This is impossible. For every $n>1$, we have gcd$(n,n-1)=1$. Since there is a prime dividing $n-1$, the result follows.



        Edit : I should have assumed $n>2$ since I need a prime divisor of $n-1$.






        share|cite|improve this answer











        $endgroup$












        • $begingroup$
          Well, $n=2$ actually works ...
          $endgroup$
          – Henning Makholm
          8 hours ago










        • $begingroup$
          @HenningMakholm you are correct, in $n=2$ case, $n-1$ is not divisible by any prime. I should have assumed that $n>2$.
          $endgroup$
          – Levent
          8 hours ago










        • $begingroup$
          what if $n-1$ and $n+1$ are twin primes, I never said $n$ had to be prime?
          $endgroup$
          – spydragon
          8 hours ago











        • $begingroup$
          @spydragon I did not assume that $n$ is a prime. Let $n>2$ be any integer. Then gcd$(n,n-1)=1$ (why?). Now $n-1>1$ so there exists a prime divisor $p$ of $n-1$ (it can be $n-1$ itself). Then $p<n$ and $p$ does not divide $n$ (since $p$ divides $n-1$ and gcd$(n,n-1)=1$).
          $endgroup$
          – Levent
          8 hours ago










        • $begingroup$
          @spydragon you want $n$ to contain every prime factor less than or equal to $n$ - it doesn't have to be prime. But the factors of $n-1$ don't divide $n$. If $n-1$ has a non-trivial prime factor, $n$ cannot contain it, Please think about this - try out some examples. $n=2$ is the only solution.
          $endgroup$
          – Mark Bennet
          8 hours ago













        10












        10








        10





        $begingroup$

        This is impossible. For every $n>1$, we have gcd$(n,n-1)=1$. Since there is a prime dividing $n-1$, the result follows.



        Edit : I should have assumed $n>2$ since I need a prime divisor of $n-1$.






        share|cite|improve this answer











        $endgroup$



        This is impossible. For every $n>1$, we have gcd$(n,n-1)=1$. Since there is a prime dividing $n-1$, the result follows.



        Edit : I should have assumed $n>2$ since I need a prime divisor of $n-1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 8 hours ago

























        answered 8 hours ago









        LeventLevent

        2,95310 silver badges26 bronze badges




        2,95310 silver badges26 bronze badges











        • $begingroup$
          Well, $n=2$ actually works ...
          $endgroup$
          – Henning Makholm
          8 hours ago










        • $begingroup$
          @HenningMakholm you are correct, in $n=2$ case, $n-1$ is not divisible by any prime. I should have assumed that $n>2$.
          $endgroup$
          – Levent
          8 hours ago










        • $begingroup$
          what if $n-1$ and $n+1$ are twin primes, I never said $n$ had to be prime?
          $endgroup$
          – spydragon
          8 hours ago











        • $begingroup$
          @spydragon I did not assume that $n$ is a prime. Let $n>2$ be any integer. Then gcd$(n,n-1)=1$ (why?). Now $n-1>1$ so there exists a prime divisor $p$ of $n-1$ (it can be $n-1$ itself). Then $p<n$ and $p$ does not divide $n$ (since $p$ divides $n-1$ and gcd$(n,n-1)=1$).
          $endgroup$
          – Levent
          8 hours ago










        • $begingroup$
          @spydragon you want $n$ to contain every prime factor less than or equal to $n$ - it doesn't have to be prime. But the factors of $n-1$ don't divide $n$. If $n-1$ has a non-trivial prime factor, $n$ cannot contain it, Please think about this - try out some examples. $n=2$ is the only solution.
          $endgroup$
          – Mark Bennet
          8 hours ago
















        • $begingroup$
          Well, $n=2$ actually works ...
          $endgroup$
          – Henning Makholm
          8 hours ago










        • $begingroup$
          @HenningMakholm you are correct, in $n=2$ case, $n-1$ is not divisible by any prime. I should have assumed that $n>2$.
          $endgroup$
          – Levent
          8 hours ago










        • $begingroup$
          what if $n-1$ and $n+1$ are twin primes, I never said $n$ had to be prime?
          $endgroup$
          – spydragon
          8 hours ago











        • $begingroup$
          @spydragon I did not assume that $n$ is a prime. Let $n>2$ be any integer. Then gcd$(n,n-1)=1$ (why?). Now $n-1>1$ so there exists a prime divisor $p$ of $n-1$ (it can be $n-1$ itself). Then $p<n$ and $p$ does not divide $n$ (since $p$ divides $n-1$ and gcd$(n,n-1)=1$).
          $endgroup$
          – Levent
          8 hours ago










        • $begingroup$
          @spydragon you want $n$ to contain every prime factor less than or equal to $n$ - it doesn't have to be prime. But the factors of $n-1$ don't divide $n$. If $n-1$ has a non-trivial prime factor, $n$ cannot contain it, Please think about this - try out some examples. $n=2$ is the only solution.
          $endgroup$
          – Mark Bennet
          8 hours ago















        $begingroup$
        Well, $n=2$ actually works ...
        $endgroup$
        – Henning Makholm
        8 hours ago




        $begingroup$
        Well, $n=2$ actually works ...
        $endgroup$
        – Henning Makholm
        8 hours ago












        $begingroup$
        @HenningMakholm you are correct, in $n=2$ case, $n-1$ is not divisible by any prime. I should have assumed that $n>2$.
        $endgroup$
        – Levent
        8 hours ago




        $begingroup$
        @HenningMakholm you are correct, in $n=2$ case, $n-1$ is not divisible by any prime. I should have assumed that $n>2$.
        $endgroup$
        – Levent
        8 hours ago












        $begingroup$
        what if $n-1$ and $n+1$ are twin primes, I never said $n$ had to be prime?
        $endgroup$
        – spydragon
        8 hours ago





        $begingroup$
        what if $n-1$ and $n+1$ are twin primes, I never said $n$ had to be prime?
        $endgroup$
        – spydragon
        8 hours ago













        $begingroup$
        @spydragon I did not assume that $n$ is a prime. Let $n>2$ be any integer. Then gcd$(n,n-1)=1$ (why?). Now $n-1>1$ so there exists a prime divisor $p$ of $n-1$ (it can be $n-1$ itself). Then $p<n$ and $p$ does not divide $n$ (since $p$ divides $n-1$ and gcd$(n,n-1)=1$).
        $endgroup$
        – Levent
        8 hours ago




        $begingroup$
        @spydragon I did not assume that $n$ is a prime. Let $n>2$ be any integer. Then gcd$(n,n-1)=1$ (why?). Now $n-1>1$ so there exists a prime divisor $p$ of $n-1$ (it can be $n-1$ itself). Then $p<n$ and $p$ does not divide $n$ (since $p$ divides $n-1$ and gcd$(n,n-1)=1$).
        $endgroup$
        – Levent
        8 hours ago












        $begingroup$
        @spydragon you want $n$ to contain every prime factor less than or equal to $n$ - it doesn't have to be prime. But the factors of $n-1$ don't divide $n$. If $n-1$ has a non-trivial prime factor, $n$ cannot contain it, Please think about this - try out some examples. $n=2$ is the only solution.
        $endgroup$
        – Mark Bennet
        8 hours ago




        $begingroup$
        @spydragon you want $n$ to contain every prime factor less than or equal to $n$ - it doesn't have to be prime. But the factors of $n-1$ don't divide $n$. If $n-1$ has a non-trivial prime factor, $n$ cannot contain it, Please think about this - try out some examples. $n=2$ is the only solution.
        $endgroup$
        – Mark Bennet
        8 hours ago













        6












        $begingroup$

        To make Lord Shark's comment explicit:



        Let $n$ be composite and let $p$ be its largest prime factor. Then certainly $nge 2p$. By Bertrand's postulate, there exists a prime $q$ between $p$ and $2p$, hence there exist primes below $n$ that do not divide $n$.






        share|cite|improve this answer











        $endgroup$












        • $begingroup$
          unless that prime $q$ is in between $n$ and $2p$
          $endgroup$
          – spydragon
          8 hours ago










        • $begingroup$
          @spydragon $p$ was chosen as the largest prime divisor of $n$ so any larger prime - e.g. the $q$ that comes from Bertrand, or any other prime larger than $p$, does not divide $n$.
          $endgroup$
          – Ethan Bolker
          3 hours ago















        6












        $begingroup$

        To make Lord Shark's comment explicit:



        Let $n$ be composite and let $p$ be its largest prime factor. Then certainly $nge 2p$. By Bertrand's postulate, there exists a prime $q$ between $p$ and $2p$, hence there exist primes below $n$ that do not divide $n$.






        share|cite|improve this answer











        $endgroup$












        • $begingroup$
          unless that prime $q$ is in between $n$ and $2p$
          $endgroup$
          – spydragon
          8 hours ago










        • $begingroup$
          @spydragon $p$ was chosen as the largest prime divisor of $n$ so any larger prime - e.g. the $q$ that comes from Bertrand, or any other prime larger than $p$, does not divide $n$.
          $endgroup$
          – Ethan Bolker
          3 hours ago













        6












        6








        6





        $begingroup$

        To make Lord Shark's comment explicit:



        Let $n$ be composite and let $p$ be its largest prime factor. Then certainly $nge 2p$. By Bertrand's postulate, there exists a prime $q$ between $p$ and $2p$, hence there exist primes below $n$ that do not divide $n$.






        share|cite|improve this answer











        $endgroup$



        To make Lord Shark's comment explicit:



        Let $n$ be composite and let $p$ be its largest prime factor. Then certainly $nge 2p$. By Bertrand's postulate, there exists a prime $q$ between $p$ and $2p$, hence there exist primes below $n$ that do not divide $n$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        answered 8 hours ago


























        community wiki





        Hagen von Eitzen












        • $begingroup$
          unless that prime $q$ is in between $n$ and $2p$
          $endgroup$
          – spydragon
          8 hours ago










        • $begingroup$
          @spydragon $p$ was chosen as the largest prime divisor of $n$ so any larger prime - e.g. the $q$ that comes from Bertrand, or any other prime larger than $p$, does not divide $n$.
          $endgroup$
          – Ethan Bolker
          3 hours ago
















        • $begingroup$
          unless that prime $q$ is in between $n$ and $2p$
          $endgroup$
          – spydragon
          8 hours ago










        • $begingroup$
          @spydragon $p$ was chosen as the largest prime divisor of $n$ so any larger prime - e.g. the $q$ that comes from Bertrand, or any other prime larger than $p$, does not divide $n$.
          $endgroup$
          – Ethan Bolker
          3 hours ago















        $begingroup$
        unless that prime $q$ is in between $n$ and $2p$
        $endgroup$
        – spydragon
        8 hours ago




        $begingroup$
        unless that prime $q$ is in between $n$ and $2p$
        $endgroup$
        – spydragon
        8 hours ago












        $begingroup$
        @spydragon $p$ was chosen as the largest prime divisor of $n$ so any larger prime - e.g. the $q$ that comes from Bertrand, or any other prime larger than $p$, does not divide $n$.
        $endgroup$
        – Ethan Bolker
        3 hours ago




        $begingroup$
        @spydragon $p$ was chosen as the largest prime divisor of $n$ so any larger prime - e.g. the $q$ that comes from Bertrand, or any other prime larger than $p$, does not divide $n$.
        $endgroup$
        – Ethan Bolker
        3 hours ago











        1












        $begingroup$

        $n=2cdotnover 2$ at most for the factoring to work out. But, by that very logic, applying it with Bertrand's Postulate $$exists cinmathbbP,nover 2leq cleq nimplies cmid n land 2nmid n$$ contradicting the possibility of an n with the property of divisibility by 2 working. which then says n doesn't include all primes less than it.






        share|cite|improve this answer











        $endgroup$

















          1












          $begingroup$

          $n=2cdotnover 2$ at most for the factoring to work out. But, by that very logic, applying it with Bertrand's Postulate $$exists cinmathbbP,nover 2leq cleq nimplies cmid n land 2nmid n$$ contradicting the possibility of an n with the property of divisibility by 2 working. which then says n doesn't include all primes less than it.






          share|cite|improve this answer











          $endgroup$















            1












            1








            1





            $begingroup$

            $n=2cdotnover 2$ at most for the factoring to work out. But, by that very logic, applying it with Bertrand's Postulate $$exists cinmathbbP,nover 2leq cleq nimplies cmid n land 2nmid n$$ contradicting the possibility of an n with the property of divisibility by 2 working. which then says n doesn't include all primes less than it.






            share|cite|improve this answer











            $endgroup$



            $n=2cdotnover 2$ at most for the factoring to work out. But, by that very logic, applying it with Bertrand's Postulate $$exists cinmathbbP,nover 2leq cleq nimplies cmid n land 2nmid n$$ contradicting the possibility of an n with the property of divisibility by 2 working. which then says n doesn't include all primes less than it.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 4 hours ago









            mlchristians

            17110 bronze badges




            17110 bronze badges










            answered 7 hours ago









            Roddy MacPheeRoddy MacPhee

            1,7432 gold badges2 silver badges24 bronze badges




            1,7432 gold badges2 silver badges24 bronze badges





















                1












                $begingroup$

                Look at it from the other side, the "half-primorials" (see http://oeis.org/A070826) of which the first composite one is 15, which as you know is $3 times 5$.



                But it's not divisible by 7, 11 or 13. Multiply those in and you get 15015. But that's not divisible by 17, 19, 23, 29, ..., 14983 or 15013. Multiply those in... actually, don't, you might tie up your computer for too long.






                share|cite|improve this answer









                $endgroup$

















                  1












                  $begingroup$

                  Look at it from the other side, the "half-primorials" (see http://oeis.org/A070826) of which the first composite one is 15, which as you know is $3 times 5$.



                  But it's not divisible by 7, 11 or 13. Multiply those in and you get 15015. But that's not divisible by 17, 19, 23, 29, ..., 14983 or 15013. Multiply those in... actually, don't, you might tie up your computer for too long.






                  share|cite|improve this answer









                  $endgroup$















                    1












                    1








                    1





                    $begingroup$

                    Look at it from the other side, the "half-primorials" (see http://oeis.org/A070826) of which the first composite one is 15, which as you know is $3 times 5$.



                    But it's not divisible by 7, 11 or 13. Multiply those in and you get 15015. But that's not divisible by 17, 19, 23, 29, ..., 14983 or 15013. Multiply those in... actually, don't, you might tie up your computer for too long.






                    share|cite|improve this answer









                    $endgroup$



                    Look at it from the other side, the "half-primorials" (see http://oeis.org/A070826) of which the first composite one is 15, which as you know is $3 times 5$.



                    But it's not divisible by 7, 11 or 13. Multiply those in and you get 15015. But that's not divisible by 17, 19, 23, 29, ..., 14983 or 15013. Multiply those in... actually, don't, you might tie up your computer for too long.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 3 hours ago









                    Robert SoupeRobert Soupe

                    12k2 gold badges21 silver badges53 bronze badges




                    12k2 gold badges21 silver badges53 bronze badges





















                        0












                        $begingroup$

                        A small dilation on the answer of Levent: Let us imagine there is such a number and there are $n$ primes smaller than that number. Then the smallest number that contains all of those primes is the product of all of those primes, which is called a primorial and is denoted $p_n#$.



                        Your question in effect asks, "Is it ever the case that $p_n+1>p_n#$?"



                        We can start by considering the number $p_n#+1$. Since every one of the first $n$ primes divides $p_n#$, and none of them divide $1$, we conclude none of them can divide $p_n#+1$. So that number must either be a prime or have prime factors. If it has prime factors, they must be other than any of the first $p_n$, and those factors must be smaller than $p_n#$, and larger than $p_n$ itself. In this case, the prime next larger than $p_n$ would have to be smaller than $p_n#$ and the answer to your question would be "No". So that forces us to consider that $p_n#+1$ is in fact a prime, and postulate that it is the prime next larger than $p_n$, i.e. $p_n+1$. If it were ever the case that $p_n+1=p_n#+1$, that would be the only possible way to answer your question in the positive.



                        Here is how Levent's answer sinks that ship. We must now consider the number $p_n#-1$. As in the case of $p_n#+1$, each $p_n$ divides $p_n#$ but none of them divide $1$, so $p_n#-1$ must either be a prime or have prime factors, in either instance smaller than $p_n#$. That being the case, $p_n#+1$ cannot be the prime next larger than $p_n$.



                        Levent's answer stands on its own in answering your question in the negative. However, the excursion taken here shows that Levent's answer also undercuts the only reasonable candidate for the kind of number you postulate.






                        share|cite|improve this answer









                        $endgroup$

















                          0












                          $begingroup$

                          A small dilation on the answer of Levent: Let us imagine there is such a number and there are $n$ primes smaller than that number. Then the smallest number that contains all of those primes is the product of all of those primes, which is called a primorial and is denoted $p_n#$.



                          Your question in effect asks, "Is it ever the case that $p_n+1>p_n#$?"



                          We can start by considering the number $p_n#+1$. Since every one of the first $n$ primes divides $p_n#$, and none of them divide $1$, we conclude none of them can divide $p_n#+1$. So that number must either be a prime or have prime factors. If it has prime factors, they must be other than any of the first $p_n$, and those factors must be smaller than $p_n#$, and larger than $p_n$ itself. In this case, the prime next larger than $p_n$ would have to be smaller than $p_n#$ and the answer to your question would be "No". So that forces us to consider that $p_n#+1$ is in fact a prime, and postulate that it is the prime next larger than $p_n$, i.e. $p_n+1$. If it were ever the case that $p_n+1=p_n#+1$, that would be the only possible way to answer your question in the positive.



                          Here is how Levent's answer sinks that ship. We must now consider the number $p_n#-1$. As in the case of $p_n#+1$, each $p_n$ divides $p_n#$ but none of them divide $1$, so $p_n#-1$ must either be a prime or have prime factors, in either instance smaller than $p_n#$. That being the case, $p_n#+1$ cannot be the prime next larger than $p_n$.



                          Levent's answer stands on its own in answering your question in the negative. However, the excursion taken here shows that Levent's answer also undercuts the only reasonable candidate for the kind of number you postulate.






                          share|cite|improve this answer









                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            A small dilation on the answer of Levent: Let us imagine there is such a number and there are $n$ primes smaller than that number. Then the smallest number that contains all of those primes is the product of all of those primes, which is called a primorial and is denoted $p_n#$.



                            Your question in effect asks, "Is it ever the case that $p_n+1>p_n#$?"



                            We can start by considering the number $p_n#+1$. Since every one of the first $n$ primes divides $p_n#$, and none of them divide $1$, we conclude none of them can divide $p_n#+1$. So that number must either be a prime or have prime factors. If it has prime factors, they must be other than any of the first $p_n$, and those factors must be smaller than $p_n#$, and larger than $p_n$ itself. In this case, the prime next larger than $p_n$ would have to be smaller than $p_n#$ and the answer to your question would be "No". So that forces us to consider that $p_n#+1$ is in fact a prime, and postulate that it is the prime next larger than $p_n$, i.e. $p_n+1$. If it were ever the case that $p_n+1=p_n#+1$, that would be the only possible way to answer your question in the positive.



                            Here is how Levent's answer sinks that ship. We must now consider the number $p_n#-1$. As in the case of $p_n#+1$, each $p_n$ divides $p_n#$ but none of them divide $1$, so $p_n#-1$ must either be a prime or have prime factors, in either instance smaller than $p_n#$. That being the case, $p_n#+1$ cannot be the prime next larger than $p_n$.



                            Levent's answer stands on its own in answering your question in the negative. However, the excursion taken here shows that Levent's answer also undercuts the only reasonable candidate for the kind of number you postulate.






                            share|cite|improve this answer









                            $endgroup$



                            A small dilation on the answer of Levent: Let us imagine there is such a number and there are $n$ primes smaller than that number. Then the smallest number that contains all of those primes is the product of all of those primes, which is called a primorial and is denoted $p_n#$.



                            Your question in effect asks, "Is it ever the case that $p_n+1>p_n#$?"



                            We can start by considering the number $p_n#+1$. Since every one of the first $n$ primes divides $p_n#$, and none of them divide $1$, we conclude none of them can divide $p_n#+1$. So that number must either be a prime or have prime factors. If it has prime factors, they must be other than any of the first $p_n$, and those factors must be smaller than $p_n#$, and larger than $p_n$ itself. In this case, the prime next larger than $p_n$ would have to be smaller than $p_n#$ and the answer to your question would be "No". So that forces us to consider that $p_n#+1$ is in fact a prime, and postulate that it is the prime next larger than $p_n$, i.e. $p_n+1$. If it were ever the case that $p_n+1=p_n#+1$, that would be the only possible way to answer your question in the positive.



                            Here is how Levent's answer sinks that ship. We must now consider the number $p_n#-1$. As in the case of $p_n#+1$, each $p_n$ divides $p_n#$ but none of them divide $1$, so $p_n#-1$ must either be a prime or have prime factors, in either instance smaller than $p_n#$. That being the case, $p_n#+1$ cannot be the prime next larger than $p_n$.



                            Levent's answer stands on its own in answering your question in the negative. However, the excursion taken here shows that Levent's answer also undercuts the only reasonable candidate for the kind of number you postulate.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 hours ago









                            Keith BackmanKeith Backman

                            1,7721 gold badge9 silver badges13 bronze badges




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Ethnic groups”„Jews, by country of origin and age”„Minority Communities in Israel: Background & Overview”„Israel”„Language in Israel”„Selected Data from the 2011 Social Survey on Mastery of the Hebrew Language and Usage of Languages”„Religions”„5 facts about Israeli Druze, a unique religious and ethnic group”„Israël”Israel Country Study Guide„Haredi city in Negev – blessing or curse?”„New town Harish harbors hopes of being more than another Pleasantville”„List of localities, in alphabetical order”„Muncitorii români, doriți în Israel”„Prietenia româno-israeliană la nevoie se cunoaște”„The Higher Education System in Israel”„Middle East”„Academic Ranking of World Universities 2016”„Israel”„Israel”„Jewish Nobel Prize Winners”„All Nobel Prizes in Literature”„All Nobel Peace Prizes”„All Prizes in Economic Sciences”„All Nobel Prizes in Chemistry”„List of Fields Medallists”„Sakharov Prize”„Țara care și-a sfidat "destinul" și se bate umăr la umăr cu Silicon Valley”„Apple's R&D center in Israel grew to about 800 employees”„Tim Cook: Apple's Herzliya R&D center second-largest in world”„Lecții de economie de la Israel”„Land use”Israel Investment and Business GuideA Country Study: IsraelCentral Bureau of StatisticsFlorin Diaconu, „Kadima: Flexibilitate și pragmatism, dar nici un compromis în chestiuni vitale", în Revista Institutului Diplomatic Român, anul I, numărul I, semestrul I, 2006, pp. 71-72Florin Diaconu, „Likud: Dreapta israeliană constant opusă retrocedării teritoriilor cureite prin luptă în 1967", în Revista Institutului Diplomatic Român, anul I, numărul I, semestrul I, 2006, pp. 73-74MassadaIsraelul a crescut in 50 de ani cât alte state intr-un mileniuIsrael Government PortalIsraelIsraelIsraelmmmmmXX451232cb118646298(data)4027808-634110000 0004 0372 0767n7900328503691455-bb46-37e3-91d2-cb064a35ffcc1003570400564274ge1294033523775214929302638955X146498911146498911

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