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Subject: "Paired sums using two sets of consecutive integers"     Previous Topic | Next Topic
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ramsey2879
guest
Sep-09-06, 08:14 AM (EST)
 
"Paired sums using two sets of consecutive integers"
 
   Let A(n) = {1,2,3,4, 8} and B(n) = {1,2,3,4, . 16} be two sets of
consecutive integers with no repetition.

Divide the set of elements B(n) into two subsets C(n) and D(n) each
comprising 8 different integers and such that every element of B(n)
is used with the condition that 8 equations of the form A(i) + C(i) =
D(i) can be simultaneously formed without using any element of A(n), C
(n) and D(n) more than once. For bookkeeping purposes i goes from 1
to 8 and matches the value of A(i) but the C(i) and D(i) are not so
ordered.

There are 48 unique solutions each involving the additional
limitation that C(1) = 1 and D(1) = 2. Thus the equation for A(1) is
1+1=2 with each of these 48 solutions. But what I am looking for
from you goes still further and there is only one such solution. The
solution which I am looking for also has C(1) = 1 and D(1) = 2 but
furthermore has the unique quality that a simple manipulation
converts the solution to one for A(n) equal {1,2,3,4 ... 9} and B(n)
= {1,2,3,4, ... 18}! This is done as follows: A(1) to A(8) and the
first half of C(n) (C(1) to C(4)) are not changed. A(9) = 9. Old C(5)
to C(8) (The second half) are each respectively promoted to become
new C(6) to C(9). New C(5)= a number between 1 and 16 that was
origionally move from B(n) to set D(n). The new values for D(n) are
determined by these conditions and comprise the remaining numbers
from the set {1,2,3 ... 18} (that are not used as part of new C(n)).
I ask this for A(n) having the starting numbers 1-8 but I can
illustrate the problem with A(n) = {1-4} since A(n) can be any
multiple of 4 even with this added complexity. It is possible to
prove that solutions to the general problem are possible only for n =
4m or for n = 4m +1. After noticing a similarity between a solution
for A(n) = 1 to 4 and another solution for A(n) = 1 to 5. I checked
to see if the pattern could be repeated for higher values of m and It
proved true. Using the pattern as a guide, I solved this problem for A
(n) = {1,2,3, ... 4m} where m = 2-4 with only paper and a pencil.
Before noting this pattern, I needed a computer to find solutions for
m > 2, and it took my program many hours to find a solution for m =
4. Now it is possible to find solutions easily for still higher
values of m. Look at my example for m = 1. See if you can spot a pattern. If not try other ways to solve the stated problem. If you solve the problem above you should be able to see the pattern and solve for m = 3 and 4 also. I have already solve it for m = 5 and 6 without having to write any computer algorithm.

Solution for m = 1
1+1 = 2
2+4 = 6
3+5 = 8
4+3 = 7

Here the number 5 is added to A(n) and 9,10 added to D(n). In
this case the new number C(3) = old D(4) and the second half of old C
(n) are promoted. The new sets are.

1 plus 1 = 2
2 plus 4 = 6
3 plus 7 = 10
4 plus 5 = 9
5 plus 3 = 8


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Ramsey2879
guest
Sep-17-06, 06:48 AM (EST)
 
1. "RE: Paired sums using two sets of consecutive integers"
In response to message #0
 
   Here is the solution set for m = 2 to 6, only I substituted letters
for like even numbers. Can anyone solve this matrix for the complete solution sets? Note that the odd number pattern gives 3/4 of the solution.

solution matrix for m = 2 to 6:

16. 1 A 3 B 11 9 7 5
18. 1 A 3 B 13 11 9 7 5

24. 1 C 3 B 5 D 17 15 13 11 9 7
26. 1 C 3 B 5 D 19 17 15 13 11 9 7

32. 1 E 3 B 5 D 7 F 23 21 19 17 15 13 11 9
34. 1 E 3 B 5 D 7 F 25 23 21 19 17 15 13 11 9

40. 1 G 3 B 5 E 7 F 9 D 29 27 25 23 21 19 17 15 13 11
42. 1 G 3 B 5 E 7 F 9 D 31 29 27 25 23 21 19 17 15 13 11

48. 1 H 3 B 5 G 7 D 9 F 11 J 35 33 31 29 27 25 23 21 19 17 15 13
50. 1 H 3 B 5 G 7 D 9 F 11 J 37 35 33 31 29 27 25 23 21 19 17 15 13



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