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

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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 ==
{{solution}}
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=== Solution 1 (official solution) ===
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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.
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[[Image:2005_USAMO_1.png‎]]
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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.
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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>.
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=== Solution 2 ===
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The proof that no arrangement exists for <math>n=pq</math>, where <math>p,q</math> are distinct primes follows from above. Apply induction to prove all other cases are possible
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Base case:
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#<math>n=p^r</math>, where <math>p</math> is a prime and <math>r</math> is a positive integer. Any arrangement suffices
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#<math>n=pqr</math>, where <math>p,q,r</math> are distinct primes. The following configuration works
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<cmath>p,pq,pr,r,qr,q,pqr</cmath>
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Inductive step: Suppose the desired arrangement exists for a composite <math>n</math>, show the arrangement exists for <math>np^r</math>, where <math>p</math> is a prime relatively prime to <math>n</math> and <math>r</math> is a positive integer
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Let <math>a_1,a_2,\cdots,a_m</math> be the arrangement of divisors of <math>n</math>, then <math>(a_i,a_{i+1})>1</math> for <math>i=1,2,\cdots,m</math>, where <math>a_{m+1}=a_1</math>. The divisors of <math>np^r</math> greater than 1 are of the form
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<cmath>a_ip^j,p^j\qquad 1\leq i\leq m,1\leq j\leq r</cmath>
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The following sequence works
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<cmath>a_1,\cdots,a_{m-1}, a_{m-1}p,\text{other divisors in arbitrary order},a_mp,a_m</cmath>
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since all other divisors are divisible by <math>p</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]]
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{{MAA Notice}}

Latest revision as of 12:40, 4 July 2013

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.

2005 USAMO 1.png

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 $n=pqr$.


Solution 2

The proof that no arrangement exists for $n=pq$, where $p,q$ are distinct primes follows from above. Apply induction to prove all other cases are possible

Base case:

  1. $n=p^r$, where $p$ is a prime and $r$ is a positive integer. Any arrangement suffices
  2. $n=pqr$, where $p,q,r$ are distinct primes. The following configuration works

\[p,pq,pr,r,qr,q,pqr\] Inductive step: Suppose the desired arrangement exists for a composite $n$, show the arrangement exists for $np^r$, where $p$ is a prime relatively prime to $n$ and $r$ is a positive integer

Let $a_1,a_2,\cdots,a_m$ be the arrangement of divisors of $n$, then $(a_i,a_{i+1})>1$ for $i=1,2,\cdots,m$, where $a_{m+1}=a_1$. The divisors of $np^r$ greater than 1 are of the form \[a_ip^j,p^j\qquad 1\leq i\leq m,1\leq j\leq r\] The following sequence works \[a_1,\cdots,a_{m-1}, a_{m-1}p,\text{other divisors in arbitrary order},a_mp,a_m\] since all other divisors are divisible by $p$.

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
1 2 3 4 5 6
All USAMO Problems and Solutions

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