Difference between revisions of "2019 USAMO Problems/Problem 1"
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<math>\implies p=q</math> since <math>p,q>0</math>. | <math>\implies p=q</math> since <math>p,q>0</math>. | ||
− | Therefore, <math>f</math> is injective. | + | Therefore, <math>f</math> is injective. It follows that <math>f^r</math> is also injective. |
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Proof: | Proof: | ||
− | <math>f^r(b)=a=f^r(a)</math> which implies <math>b=a</math> by | + | <math>f^r(b)=a=f^r(a)</math> which implies <math>b=a</math> by injectivity of <math>f^r</math>. |
Revision as of 15:13, 5 May 2019
Problem
Let be the set of positive integers. A function
satisfies the equation
for all positive integers
. Given this information, determine all possible values of
.
Solution
Let denote the result when
is applied to
times.
If
, then
and
since
.
Therefore, is injective. It follows that
is also injective.
Lemma 1: If and
, then
.
Proof:
which implies
by injectivity of
.
Lemma 2: If , and
is odd, then
.
Proof:
Let . Since
,
. So,
.
.
Since ,
This proves Lemma 2.
I claim that for all odd
.
Otherwise, let be the least counterexample.
Since , either
, contradicted by Lemma 1 since
is odd and
.
, also contradicted by Lemma 1 by similar logic.
and
, which implies that
by Lemma 2.
This proves the claim.
By injectivity, is not odd.
I will prove that
can be any even number,
. Let
, and
for all other
. If
is equal to neither
nor
, then
. This satisfies the given property.
If is equal to
or
, then
since
is even and
. This satisfies the given property.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2019 USAMO (Problems • Resources) | ||
First Problem | Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |