1984 USAMO Problems/Problem 2
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 , 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
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 |
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