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. | + | (''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 1 (official solution) === |
+ | 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: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. | ||
+ | |||
+ | ''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>. | ||
+ | |||
+ | |||
+ | === Solution 2 === | ||
+ | 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 | ||
+ | |||
+ | Base case: | ||
+ | #<math>n=p^r</math>, where <math>p</math> is a prime and <math>r</math> is a positive integer. Any arrangement suffices | ||
+ | #<math>n=pqr</math>, where <math>p,q,r</math> are distinct primes. The following configuration works | ||
+ | <cmath>p,pq,pr,r,qr,q,pqr</cmath> | ||
+ | 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 | ||
+ | |||
+ | 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 | ||
+ | <cmath>a_ip^j,p^j\qquad 1\leq i\leq m,1\leq j\leq r</cmath> | ||
+ | The following sequence works | ||
+ | <cmath>a_1,\cdots,a_{m-1}, a_{m-1}p,\text{other divisors in arbitrary order},a_mp,a_m</cmath> | ||
+ | since all other divisors are divisible by <math>p</math>. | ||
+ | |||
+ | {{alternate solutions}} | ||
== See also == | == See also == | ||
+ | * <url>viewtopic.php?t=34314 Discussion on AoPS/MathLinks</url> | ||
+ | |||
{{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]] | ||
+ | {{MAA Notice}} |
Latest revision as of 12:40, 4 July 2013
Problem
(Zuming Feng) Determine all composite positive integers for which it is possible to arrange all divisors of 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 , where and are distinct primes. In that case, the numbers to be arranged are ; and , and in any circular arrangement, and will be adjacent. We claim that the desired circular arrangement exists in all other cases. If where , an arbitrary circular arrangement works. Henceforth we assume that has prime factorization , where and either or else . To construct the desired circular arrangement of , start with the circular arrangement of as shown.
Then between and , place (in arbitrary order) all other members of that have as their smallest prime factor. Between and , place all members of other than that have as their smallest prime factor. Continue in this way, ending by placing between and . It is easy to see that each element of 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 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 and .
Solution 2
The proof that no arrangement exists for , where are distinct primes follows from above. Apply induction to prove all other cases are possible
Base case:
- , where is a prime and is a positive integer. Any arrangement suffices
- , where are distinct primes. The following configuration works
Inductive step: Suppose the desired arrangement exists for a composite , show the arrangement exists for , where is a prime relatively prime to and is a positive integer
Let be the arrangement of divisors of , then for , where . The divisors of greater than 1 are of the form The following sequence works since all other divisors are divisible by .
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 (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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