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Difference between revisions of "1984 USAMO Problems/Problem 2"

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== Problem ==
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The geometric mean of any set of <math>m</math> non-negative numbers is the <math>m</math>-th root of their product.
  
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<math>\quad (\text{i})\quad</math> For which positive integers <math>n</math> is there a finite set <math>S_n</math> of <math>n</math> distinct positive integers such that the geometric mean of any subset of <math>S_n</math> is an integer?
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<math>\quad (\text{ii})\quad</math> Is there an infinite set <math>S</math> of distinct positive integers such that the geometric mean of any finite subset of <math>S</math> is an integer?
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==Solution==
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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)
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== See Also ==
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{{USAMO box|year=1984|num-b=1|num-a=3}}
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{{MAA Notice}}
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[[Category:Olympiad Number Theory Problems]]

Revision as of 00:05, 20 April 2021

Problem

The geometric mean of any set of $m$ non-negative numbers is the $m$-th root of their product.

$\quad (\text{i})\quad$ For which positive integers $n$ is there a finite set $S_n$ of $n$ distinct positive integers such that the geometric mean of any subset of $S_n$ is an integer?

$\quad (\text{ii})\quad$ Is there an infinite set $S$ of distinct positive integers such that the geometric mean of any finite subset of $S$ is an integer?

Solution

a) We claim that for any numbers $p_1$, $p_2$, ... $p_n$, $p_1^{n!}, p_2^{n!}, ... p_n^{n!}$ will satisfy the condition, which holds for any number $n$.

Since $\sqrt[n] ab = \sqrt[n] a * \sqrt[n] b$, we can separate each geometric mean into the product of parts, where each part is the $k$th root of each member of the subset and the subset has $k$ members.

Assume our subset has $k$ members. Then, we know that the $k$th root of each of these members is an integer (namely $p^{n!/k}$), because $k \leq n$ and thus $k | n!$. Since each root is an integer, the geometric mean will also be an integer.

b) If we define $q$ as an arbitrarily large number, and $x$ and $y$ as numbers in set $S$, we know that ${\sqrt[q]{\frac{x}{y}}}$ is irrational for large enough $q$, meaning that it cannot be expressed as the fraction of two integers. However, both the geometric mean of the set of $x$ and $q-1$ other arbitrary numbers in $S$ and the set of $y$ and the same other $q-1$ numbers are integers, so since the other numbers cancel out, the geometric means divided, or ${\sqrt[q]{\frac{x}{y}}}$, must be rational. This is a contradiction, so no such infinite $S$ 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

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