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Relative primes in sums of sets

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  • Kermit Rose
    3. Some clarification? Posted by: chrisdarroch chrisdarr2@hotmail.com chrisdarroch Date: Fri Jun 6, 2008 10:23 pm ((PDT)) Just an additional explanation of
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      3. Some clarification?
      Posted by: "chrisdarroch" chrisdarr2@... chrisdarroch
      Date: Fri Jun 6, 2008 10:23 pm ((PDT))

      Just an additional explanation of what I am asking.

      Consider a value n = 38.

      Consider values x =[13,17,19,23,29,31,37,41,43,47] which are all
      indivible(as it were) by any member of A.

      ***************

      Kermit says:

      Where A = [2,3,5,7,11]

      ************
      chrisdarroch says:

      Consider the results of n+x: [51,55,57,61,75,79,81,85]

      *******

      Kermit says:


      38 + [13,17,19,23,29,31,37,41,43,47]
      =[51,55,57,61, 75, 79, 81, 85]





      Since A = [2,3,5,7,11], consider representing each of the numbers in x ,

      [13,17,19,23,29,31,37,41,43,47]
      by the set of remainders when divided by an element in A.

      That is,

      13 = [1,1,3,6,2] since 13 has remainder 1 when divided by 2, remainder 1 when divided by 3,
      remainder 3 when divided by 5, remainder 6 when divided by 7, remainder 2 when divided by 11.

      Make a matrix from x using the remainders when divided by elements of A.

      A= 2 3 5 7 11
      13 1 1 1 6 2
      17 1 2 2 3 6
      19 1 1 4 5 8
      23 1 2 3 2 1
      29 1 2 4 1 7
      31 1 1 1 3 9
      37 1 1 2 2 4
      41 1 2 1 6 8
      43 1 1 3 1 10
      47 1 2 2 5 3

      By looking at the column under the 2, we see that if n is odd, every sum will be divisible by 2,
      and if n is even, no sum will be divisible by 2.

      By looking at the column under the 3, we see that if n is a multiple of 3, that no sum will be divisible by 3,
      and that if n has remainder 1 when divided by 3, that the number of 2's under the 3 is the number of sums that
      will be divisible by 3.
      Etc
      Since under 3 there are 1 and 2,
      and under 5 there are 1,2,3,4
      and under 7 there are 1,2,3,5,6, [ but not 4]

      and under 11 there are 1,2,3,4,6,7,8,9,10, [ but not 5]



      then we can easily calculate what n to use to make every sum not divisible by any element of A.

      n must be divisible by 2.
      n must be divisible by 3.
      n must be divisible by 5.
      n must either be divisible by 7 or have remainder 3 when divided by 7.
      n must either be divisible by 11, or have remainder 6 when divided by 11.

      Thus n must be one of the four forms.

      n1 = 2 * 3 * 5 * 7 * 11 k = 2310 k
      n2 = 2310 k + 1260
      n3 = 2310 k + 990
      n4 = 2310 k + 2250

      If n is any of these four forms, then every sum will be not divisible by
      any element of A.





      ********

      chrisdarroch says:

      All of these results are divisible by at least one member of
      A:[2,3,5,7,11] except for 61 and 79 which are primes. The largest gap
      within which I have found a prime thus far is 10 lets say, i.e.
      between 51 and the first prime of 61.


      I might find a relative prime instead of a prime and that gap would
      count for example if I added x = 131 to n = 38 I would get 169 which
      is divisible by 13 but not any member of A and this is how I define
      relatively prime in this case.



      Chris

      I cannot see that tautology here unless I am using the phrase
      Relatively Prime improperly.


      *******

      Kermit says:

      Predictions of gaps can also be determined by looking at the remainder matrix,
      but it is much more complicated.
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