Difference between revisions of "2005 USAMO Problems/Problem 1"

(Official solution (image needed))
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== Problem ==
 
== Problem ==
Determine all composite positive integers <math>n</math> for which it is possible to arrange all divisors of <math>n</math> that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.
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(''Zuming Feng'') Determine all composite positive integers <math>n</math> for which it is possible to arrange all divisors of <math>n</math> that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.
  
 
== Solution ==
 
== Solution ==
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''Note.'' In graph theory terms, this construction yields a Hamiltonian cycle in the graph with vertex set <math>D_n</math> in which two vertices form an edge if the two corresponding numbers have a common prime factor. The graphs below illustrate the construction for the special cases <math>n=p^{2}q</math> and <math>n=pqr</math>.
 
''Note.'' In graph theory terms, this construction yields a Hamiltonian cycle in the graph with vertex set <math>D_n</math> in which two vertices form an edge if the two corresponding numbers have a common prime factor. The graphs below illustrate the construction for the special cases <math>n=p^{2}q</math> and <math>n=pqr</math>.
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{{alternate solutions}}
  
 
== See also ==
 
== See also ==
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* <url>viewtopic.php?t=34314 Discussion on AoPS/MathLinks</url>
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{{USAMO newbox|year=2005|before=First Question|num-a=2}}
 
{{USAMO newbox|year=2005|before=First Question|num-a=2}}
  
 
[[Category:Olympiad Number Theory Problems]]
 
[[Category:Olympiad Number Theory Problems]]

Revision as of 13:24, 3 May 2008

Problem

(Zuming Feng) Determine all composite positive integers $n$ for which it is possible to arrange all divisors of $n$ that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.

Solution

Solution 1 (official solution)

No such circular arrangement exists for $n=pq$, where $p$ and $q$ are distinct primes. In that case, the numbers to be arranged are $p$; $q$ and $pq$, and in any circular arrangement, $p$ and $q$ will be adjacent. We claim that the desired circular arrangement exists in all other cases. If $n=p^e$ where $e\ge2$, an arbitrary circular arrangement works. Henceforth we assume that $n$ has prime factorization $p^{e_1}_{1}p^{e_2}_{2}\cdots p^{e_k}_k$, where $p_1<p_2<\cdots<p_k$ and either $k>2$ or else $\max(e1,e2)>1$. To construct the desired circular arrangement of $D_n:=\lbrace d:d|n\ \text{and}\ d>1\rbrace$, start with the circular arrangement of $n,p_{1}p_{2},p_{2}p_{3}\ldots,p_{k-1}p_{k}$ as shown.


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Then between $n$ and $p_{1}p_{2}$, place (in arbitrary order) all other members of $D_n$ that have $p_1$ as their smallest prime factor. Between $p_{1}p_{2}$ and $p_{2}p_{3}$, place all members of $D_n$ other than $p_{2}p_{3}$ that have $p_2$ as their smallest prime factor. Continue in this way, ending by placing $p_k,p^{2}_{k},\ldots,p^{e_k}_{k}$ between $p_{k-1}p_k$ and $n$. It is easy to see that each element of $D_n$ is placed exactly one time, and any two adjacent elements have a common prime factor. Hence this arrangement has the desired property.

Note. In graph theory terms, this construction yields a Hamiltonian cycle in the graph with vertex set $D_n$ in which two vertices form an edge if the two corresponding numbers have a common prime factor. The graphs below illustrate the construction for the special cases $n=p^{2}q$ and $n=pqr$.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also

  • <url>viewtopic.php?t=34314 Discussion on AoPS/MathLinks</url>
2005 USAMO (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
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All USAMO Problems and Solutions
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