PizzaPerson wrote: 19049171 I did not solve the problem, but I am pretty sure there are at most $\binom{28}{3}=3276$ ways to distribute three hats to the dwarves (since certainly no two dwarves can have the same collection of hats). So your answer can't be correct.

I have a construction. Consider every combination of numbers $a,b,c$ in the specified range such that $a+b=c$. Then have one dwarf get each possibility. It is clear that no two dwarfs can share at least two in common; for example, if two dwarfs shared $a$ and $b$, they also must have shared $c$ by the condition $a+b=c$, a contradiction. Now to count the number of triples like this. If the first number is $n$, the second number can be anything from $n+1$ to $28-n$, giving $28-2n$ possibilities. Therefore there are $26+24+\ldots+2=182$ possibilities, and thus this is a lower bound.

Okay, now I have an argument that proves $182$ is maximal. Consider creating an upper bound for the number of dwarfs with a certain middle value. If the middle number is $n$, there must be at most $n-1$ possibilities, as the lower number must be different for all dwarfs with a common middle number. Similarly, there must be at most $28-n$ possibilities, as the higher number must be different for all dwarfs with a common middle number. Therefore there can be at most $\min(n-1,28-n)$ dwarfs that have a certain middle number $n$. It is clear that the middle numbers must be between $2$ and $27$, inclusive, so summing over this interval, we get that there can be at most $1+2+3+\ldots+12+13+13+12+\ldots+3+2+1=182$ dwarfs, and we are done.

bluelinfish wrote: I have a construction. Consider every combination of numbers $a,b,c$ in the specified range such that $a+b=c$. .... . This is not correct. For example one may add the triple (26,27,28) and still satisfy conditions. bluelinfish wrote: Consider creating an upper bound for the number of dwarfs with a certain middle value. .... . No guarantee that the average value is an integer.

gogyan wrote: bluelinfish wrote: I have a construction. Consider every combination of numbers $a,b,c$ in the specified range such that $a+b=c$. .... . This is not correct. For example one may add the triple (26,27,28) and still satisfy conditions. bluelinfish wrote: Consider creating an upper bound for the number of dwarfs with a certain middle value. .... . No guarantee that the average value is an integer. his proof is right.

No no, he really means $a+b=c$. Otherwise, we would not get enough triples. Note that middle number is just $b$ if we assume that $a<b<c$. It's not the average. Of course $b$ is always an integer. We are just counting the number of triples with a fixed "middle number" $b$. gogyan wrote: This is not correct. For example one may add the triple (26,27,28) and still satisfy conditions. I don't understand. We already have the triple $(1,27,28)$ since $1+27=28$. So we cannot add your triple $(26,27,28)$...

Tintarn wrote: No no, he really means $a+b=c$. Otherwise, we would not get enough triples. In fact,this is just a different construction,which also works well,as you can check.

Answer:$182$. Example: Number $i$. is in a triplet with $\{i+1,i+2\},\{i+2,i+4\},\{i+3,i+6\}...$

Example

Proof: If a dwarf wears $a<b<c$ in the festivals, then its triplet is $(a,b,c)$. Let $A=\{1,2,3,...,14\}$ and $B=\{15,16,17,...,28\}$. A triplet has three elements so each triplet's at least two elements are in the same group. Assume that there are $183$ or more dwarfs. Then, we have $183$ triplets such that at least $92$ of them has two elements in the same group. If there are $92$ triplets which has at least $2$ elements in $A$: Look at the first $2$ elements of these $92$ triplets'. They all should be in $A$.There are $\binom{14}{2}=91$ probabilities about different pairs but there are $92$ pairs which are in $A$. There should be at least $2$ pairs which are the same so we get contradiction. If there are $92$ triplets which has at least $2$ elements in $B$: Look at the last $2$ elements of these $92$ triplets'. They all should be in $B$.There are $\binom{14}{2}=91$ probabilities about different pairs but there are $92$ pairs which are in $B$. There should be at least $2$ pairs which are the same so we get contradiction.