## Relative primes in sums of sets

Expand Messages
• 3. Some clarification? Posted by: chrisdarroch chrisdarr2@hotmail.com chrisdarroch Date: Fri Jun 6, 2008 10:23 pm ((PDT)) Just an additional explanation of
Message 1 of 1 , Jun 7, 2008
• 0 Attachment
3. Some clarification?
Posted by: "chrisdarroch" chrisdarr2@... chrisdarroch
Date: Fri Jun 6, 2008 10:23 pm ((PDT))

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.
Your message has been successfully submitted and would be delivered to recipients shortly.