Difference between revisions of "1984 USAMO Problems/Problem 2"
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==Solution== | ==Solution== | ||
| − | {{ | + | a) We claim that for any numbers <math>p_1</math>, <math>p_2</math>, ... <math>p_n</math>, <math>p_1^{n!}, p_2^{n!}, ... p_n^{n!}</math> will satisfy the condition, which holds for any number <math>n</math>. |
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| + | Since <math>\sqrt[n] ab = \sqrt[n] a * \sqrt[n] b</math>, we can separate each geometric mean into the product of parts, where each part is the <math>k</math>th root of each member of the subset and the subset has <math>k</math> members. | ||
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| + | Assume our subset has <math>k</math> members. Then, we know that the <math>k</math>th root of each of these members is an integer (namely <math>p^{n!/k}</math>), because <math>k \leq n</math> and thus <math>k | n!</math>. Since each root is an integer, the geometric mean will also be an integer. | ||
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| + | b) If we define <math>q</math> as an arbitrarily large number, and <math>x</math> and <math>y</math> as numbers in set <math>S</math>, we know that <math>{\sqrt[q]{\frac{x}{y}}}</math> is irrational for large enough <math>q</math>, meaning that it cannot be expressed as the fraction of two integers. However, both the geometric mean of the set of <math>x</math> and <math>q-1</math> other arbitrary numbers in <math>S</math> and the set of <math>y</math> and the same other <math>q-1</math> numbers are integers, so since the other numbers cancel out, the geometric means divided, or <math>{\sqrt[q]{\frac{x}{y}}}</math>, must be rational. This is a contradiction, so no such infinite <math>S</math> is possible. | ||
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| + | -aops111 (first solution dont bully me) | ||
== See Also == | == See Also == | ||
Revision as of 23:05, 19 April 2021
Problem
The geometric mean of any set of 
non-negative numbers is the -th root of their product.
For which positive integers 
is there a finite set 
of distinct positive integers such that the geometric mean of any subset of is an integer?
Is there an infinite set 
of distinct positive integers such that the geometric mean of any finite subset of is an integer?
Solution
a) We claim that for any numbers 
, 
, ... 
, 
will satisfy the condition, which holds for any number .
Since ![$\sqrt[n] ab = \sqrt[n] a * \sqrt[n] b$](http://latex.artofproblemsolving.com/e/8/d/e8d795ea6fb057679e141a7b685896f493b4681f.png)
, we can separate each geometric mean into the product of parts, where each part is the 
th root of each member of the subset and the subset has members.
Assume our subset has members. Then, we know that the th root of each of these members is an integer (namely 
), because 
and thus 
. Since each root is an integer, the geometric mean will also be an integer.
b) If we define 
as an arbitrarily large number, and 
and 
as numbers in set , we know that ![${\sqrt[q]{\frac{x}{y}}}$](http://latex.artofproblemsolving.com/a/b/5/ab5e4ffba6e1885d08ca59a4750888f387b8b6df.png)
is irrational for large enough , meaning that it cannot be expressed as the fraction of two integers. However, both the geometric mean of the set of and 
other arbitrary numbers in and the set of and the same other numbers are integers, so since the other numbers cancel out, the geometric means divided, or , must be rational. This is a contradiction, so no such infinite is possible.
-aops111 (first solution dont bully me)
See Also
| 1984 USAMO (Problems • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
