Difference between revisions of "2006 USAMO Problems/Problem 4"
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== Problem == | == Problem == | ||
− | Find all positive integers <math>n</math> such that there are <math>k\ge 2</math> positive rational numbers <math>a_1,a_2,\ldots a_k</math> satisfying <math>a_1+a_2+...+a_k=a_1\ | + | (''Ricky Liu'') Find all positive integers <math>n</math> such that there are <math>k\ge 2</math> positive rational numbers <math>a_1, a_2, \ldots, a_k</math> satisfying <math>a_1 + a_2 + \cdots + a_k = a_1\cdot a_2\cdots a_k = n</math>. |
− | == | + | |
− | == See | + | == Solutions == |
− | *[[ | + | |
+ | === Solution 1 === | ||
+ | First, consider composite numbers. We can then factor <math>n</math> into <math>p_1p_2.</math> It is easy to see that <math>p_1+p_2\le n</math>, and thus, we can add <math>(n-p_1-p_2)</math> 1s in order to achieve a sum and product of <math>n</math>. For <math>p_1+p_2=n</math>, which is only possible in one case, <math>n=4</math>, we consider <math>p_1=p_2=2</math>. | ||
+ | |||
+ | Secondly, let <math>n</math> be a prime. Then we can find the following procedure: Let <math>a_1=\frac{n}{2}, a_2=4, a_3=\frac{1}{2}</math> and let the rest of the <math>a_k</math> be 1. The only numbers we now need to check are those such that <math>\frac{n}{2}+4+\frac{1}{2}>n\Longrightarrow n<9</math>. Thus, we need to check for <math>n=1,2,3,5,7</math>. One is included because it is neither prime nor composite. | ||
+ | |||
+ | For <math>n=1</math>, consider <math>a_1a_2\hdots a_k=1</math>. Then by AM-GM, <math>a_1+a_2+\hdots+a_k\ge k\sqrt[k]{1}>1</math> for <math>k\ge 2</math>. Thus, <math>n=1</math> is impossible. | ||
+ | |||
+ | If <math>n=2</math>, once again consider <math>a_1a_2\hdots a_k=2</math>. Similar to the above, <math>a_1+a_2+\hdots\ge k\sqrt[k]{2}>2</math> for <math>k\ge 2</math> since <math>\sqrt[k]{2}>1</math> and <math>k>2</math>. Obviously, <math>n=2</math> is then impossible. | ||
+ | |||
+ | If <math>n=3</math>, let <math>a_1a_2\hdots a_k=3</math>. Again, <math>a_1+a_2+\hdots\ge k\sqrt[k]{3}>3</math>. This is obvious for <math>k\ge 3</math>. Now consider <math>k=2</math>. Then <math>2\sqrt{3}\approx 3.4</math> is obviously greater than <math>3</math>. Thus, <math>n=3</math> is impossible. | ||
+ | |||
+ | If <math>n=5</math>, proceed as above and consider <math>k=2</math>. Then <math>a_1+a_2=5</math> and <math>a_1a_2=5</math>. However, we then come to the quadratic <math>a_1^2-5a_1+5=0 \Longrightarrow a_1=\frac{5\pm\sqrt{5}}{2}</math>, which is not rational. For <math>k=3</math> and <math>k=4</math> we note that <math>\sqrt[3]{5}>\frac{5}{3}</math> and <math>\sqrt[4]{5}>\frac{5}{4}</math>. This is trivial to prove. If <math>k\ge 5</math>, it is obviously impossible, and thus <math>n=5</math> does not work. | ||
+ | |||
+ | The last case, where <math>n=7</math>, is possible using the following three numbers. <math>a_1=\frac{9}{2}, a_2=\frac{4}{3}, a_3=\frac{7}{6}</math> shows that <math>n=7</math> is possible. | ||
+ | |||
+ | Hence, <math>n</math> can be any positive integer greater than <math>3</math> with the exclusion of <math>5</math>. | ||
+ | |||
+ | {{alternate solutions}} | ||
+ | |||
+ | == See also == | ||
+ | * <url>viewtopic.php?t=84554 Discussion on AoPS/MathLinks</url> | ||
+ | |||
+ | {{USAMO newbox|year=2006|num-b=3|num-a=5}} | ||
+ | |||
+ | [[Category:Olympiad Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:43, 5 August 2014
Contents
Problem
(Ricky Liu) Find all positive integers such that there are positive rational numbers satisfying .
Solutions
Solution 1
First, consider composite numbers. We can then factor into It is easy to see that , and thus, we can add 1s in order to achieve a sum and product of . For , which is only possible in one case, , we consider .
Secondly, let be a prime. Then we can find the following procedure: Let and let the rest of the be 1. The only numbers we now need to check are those such that . Thus, we need to check for . One is included because it is neither prime nor composite.
For , consider . Then by AM-GM, for . Thus, is impossible.
If , once again consider . Similar to the above, for since and . Obviously, is then impossible.
If , let . Again, . This is obvious for . Now consider . Then is obviously greater than . Thus, is impossible.
If , proceed as above and consider . Then and . However, we then come to the quadratic , which is not rational. For and we note that and . This is trivial to prove. If , it is obviously impossible, and thus does not work.
The last case, where , is possible using the following three numbers. shows that is possible.
Hence, can be any positive integer greater than with the exclusion of .
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=84554 Discussion on AoPS/MathLinks</url>
2006 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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