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

Revision as of 19:15, 27 July 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.

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.

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 No such circular arrangement exists for <math>n=pq</math>, where <math>p</math> and <math>q</math> are distinct primes. In that case, the numbers to be arranged are <math>p</math>; <math>q</math> and <math>pq</math>, and in any circular arrangement, <math>p</math> and <math>q</math> will be adjacent. We claim that the desired circular arrangement exists in all other cases. If <math>n=p^e</math> where <math>e\ge2</math>, an arbitrary circular arrangement works. Henceforth we assume that <math>n</math> has prime factorization <math>p^{e_1}_{1}p^{e_2}_{2}\cdots p^{e_k}_k</math>, where <math>p_1<p_2<\cdots<p_k</math> and either <math>k>2</math> or else <math>\max(e1,e2)>1</math>. To construct the desired circular arrangement of <math>D_n:=\lbrace d:d|n\ \text{and}\ d>1\rbrace</math>, start with the circular arrangement of <math>n,p_{1}p_{2},p_{2}p_{3}\ldots,p_{k-1}p_{k}</math> as shown.
-{{image}}
+[[Image:2005_USAMO_1.png‎]]
 Then between <math>n</math> and <math>p_{1}p_{2}</math>, place (in arbitrary order) all other members of <math>D_n</math> that have <math>p_1</math> as their smallest prime factor. Between <math>p_{1}p_{2}</math> and <math>p_{2}p_{3}</math>, place all members of <math>D_n</math> other than <math>p_{2}p_{3}</math> that have <math>p_2</math> as their smallest prime factor. Continue in this way, ending by placing <math>p_k,p^{2}_{k},\ldots,p^{e_k}_{k}</math> between <math>p_{k-1}p_k</math> and <math>n</math>. It is easy to see that each element of <math>D_n</math> is placed exactly one time, and any two adjacent elements have a common prime factor. Hence this arrangement has the desired property.

2005 USAMO (Problems • Resources)
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Difference between revisions of "2005 USAMO Problems/Problem 1"

Revision as of 19:15, 27 July 2008

Contents

Problem

Solution

Solution 1 (official solution)

See also