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I am trying to solve the following problem.

Assume that we are given 26 distinct positive integers. Show that either there exist 6 of them $x_1<x_2<x_3<x_4<x_5<x_6$, with $x_1$ dividing $x_2$, $x_2$ dividing $x_3$, $x_3$ dividing $x_4$, $x_4$ dividing $x_5$ and $x_5$ dividing $x_6$ or there exist six of them, such that none of them divides another one of these six.

A possibly good start is to assume that, in every six of these numbers, there exists at least one dividing another one of the same six.

Update. I have found a solution of the problem (for 17 numbers though) in a Russian site. As unbelievable as it may sound, this problem was a question in a 1983 Soviet Mathematics contest (Турниры Городов) for student of 7-8 grades!

I am presenting the solution I found in that site below as an answer, and it is generalised for $n^2+1$ distinct integers, where we show that either there exist $n+1$ of them dividing each other or none diving none else.

Yiorgos S. Smyrlis
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    This is not a duplicate of the question on increasing and decreasing subsequences, as divisibility is a partial order, not a strict linear order. – ShreevatsaR Jan 26 '14 at 09:33

3 Answers3

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Apply Dilworth's Theorem to the poset of the $26$ integers under the divisibility relation. Either there is an antichain of length $6$ (no two of $6$ integers divide one another) or the set can be partitioned into at most $5$ chains (sequences where each integer divides the next), one of which must have length at least $6$ by the pigeonhole principle.

universalset
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Define a chain as a sequence $(y_j)$ such that $y_1|y_2 |...|y_k$ and an antichain a sequence $(y_j)$ such that none of the $y_j$ divides one of the remaining $y_i, i\neq j$.

To each $x$ in the initial sequence attach a pair of numbers $(i_x,j_x)$ such that $i_x$ is the the length of the longest chain ending in $x$ and $j_x$ is the length of the longest antichain starting with $x$.

Pick now two elements $x\neq y$ such that $x$ is before $y$ in the sequence $(x_n)$. If $x | y$ then $i_x<i_y$ since any chain ending in $x$ can be extended to a longer chain ending in $y$. If $x \nmid y$ then $j_x > j_y$ since any antichain starting in $y$ can be extended to a longer antichain starting in $x$. Therefore the application $x \mapsto (i_x,j_x)$ is injective.

Consider a sequence with $mn+1$ elements, and suppose the length of the longest chain is $m$ and the length of the longest antichain is $n$. Therefore the map $x \mapsto (i_x,j_x)$ is an injection from a set with $mn+1$ elements into a set with $mn$ elements. This is a contradiction proving that there is a chain of length $m+1$ or an antichain of length $n+1$.

In the problem presented here we have $m=n=5$.

Beni Bogosel
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  • The third paragraph should end $x \mapsto (i_x,j_x)$ instead of $x \mapsto (i_x,i_y)$. (I tried editing your answer, but the change was less than 6 characters, which isn't allowed.) – Steve Kass Dec 28 '13 at 02:46
  • @SteveKass: Got it fixed – Beni Bogosel Dec 28 '13 at 10:14
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    Your definition of antichain isn't right: the elements have to be such that none divides another. $2\nmid 5$ and $5\nmid 6$ but these three don't form an antichain. – universalset Jan 25 '14 at 14:41
  • I have the same doubt as @universalset: For the desired conclusion of the question (and for the general meaning of "antichain") to hold, you must define your antichain as a sequence such that none of them divides any other. But then we can no longer say that if $x\nmid y$, then an antichain starting at $y$ can be extended to a longer antichain starting at $x$. Indeed, the conclusion of this proof, that $x\mapsto (i_x,j_x)$ is injective, is not true: Consider the sequence $(2,3,4)$. Then, the pairs corresponding to $2$ and $3$ are both $(1,2)$ and $(1,2)$. – ShreevatsaR Jan 29 '14 at 03:57
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I have found this solution (for 17 numbers though) in a Russian site, as this problem was a question in a 1983 Soviet Mathematics contest (Турниры Городов) for student of 7-8 classes!

Solution. We shall attach on each of our numbers $$ x_1<x_2<\cdots<x_{26}, $$ an index next to their subscript in the following fashion:

$x_1$ becomes $x_{1,1}$.

If $x_1$ divides $x_2$, then $x_2$ becomes $x_{2,2}$, otherwise it becomes $x_{2,1}$.

In general, if we have attached indices to $x_1,\ldots,x_k$, then the index of $x_{k+1}$ will be $1$ is none of the $x_1,\ldots,x_k$ divides $x_{k+1}$, or it will become $i+1$, if $i$ is the largest index of all the numbers which divide $x_{k+1}$.

If the index of some of the $x_j$'s is at least $6$ then we have have a sequence $$ x_{k_1,1} \mid x_{k_2,2} \mid x_{k_3,3} \mid x_{k_4,4} \mid x_{k_5,5}\mid x_{k_6,6}. $$ In all the indices are less or equal to $5$, then some number in $\{1,2,3,4,5\}$, say $j$, is necessarily the index of at least $6$ numbers, i.e. $$ x_{k_1,j} < x_{k_2,j} < x_{k_3,j} < x_{k_4,j} < x_{k_5,j} < x_{k_6,j}, $$ which means that none of the above $6$ numbers divide another one of them.

Note. Τhis can be generalised to finding a chain of $k$ numbers, fully ordered by division, or a subset of $\ell$ numbers, with none of them dividing another one, among $m$ distinct positive integers, whenever $(k-1)(\ell-1)\le m-1$.

Yiorgos S. Smyrlis
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