(In reply to
re(3): No Subject by ed bottemiller)
Hi, everybody!
I am addressing all those floobers that spent considerable amount of times trying to get the maximal value of EN. I did not realize that solving for n=20 is so much harder than, say, n=8. After private email exchange between me and Justin I became convinced that to climb to higher values necessitates solving for each new value "from the scratches" and is amazingly time-consuming.
Since I do not have the correct answer, I would like
- To tell you the history of my post.
- To define what exactly we are looking for.
Let start with the "original" text from some math contest at
high school level:
Let S be a set of 20 distinct positive integers, all less than 97.
Show that there exist four distinct elements a, b, c, d in S such that
a + b = c + d.
SOLUTION:
Is based on the pigeonhole principle: less sums than pairs are needed.
20*19*.5=190 PAIRS
SMALLEST SUM =1+2=3
LARGEST =96+95 =191
3,4,5,….191
so # of distinct sums* =191-2=189
i.e. 189 SUMS FOR 190 PAIRS
189 is less than 190
qed
This kind of question has nothing to do with the maximal value of EN ,
a value-added and challenging contribution by my humble self ,
requesting the following:
1.Provide a list of 20 integers, the lowest =1 , highest EN-1 s..t.
at least two distinct pairs satisfy a+b=c+d.
2. Prove, either analytically or by exaustive run that fror EN all
Combination of pairs for any subset of 20 members produce distinct
Sums.
If it is too tiresome and no shortcuts are known
let's solve the problem for a reduced set , say 12 numbers.
PLEASE COMMENT re: official solution.
re(4): No Subject FOR ALL SOLVERS
