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