Difference between revisions of "2019 USAJMO Problems/Problem 6"
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Assume that <math>m+n\ne 2^k</math>, so then <math>m+n\equiv 0\pmod{p}</math> for some odd prime <math>p</math>. Then <math>m\equiv -n\pmod{p}</math>, so <math>\frac{m}{n}\equiv \frac{n}{m}\equiv -1\pmod{p}</math>. We see that the arithmetic mean is <math>\frac{-1+(-1)}{2}\equiv -1\pmod{p}</math> and the harmonic mean is <math>\frac{2(-1)(-1)}{-1+(-1)}\equiv -1\pmod{p}</math>, so if 1 can be written then <math>1\equiv -1\pmod{p}</math> and <math>2\equiv 0\pmod{p}</math> which is obviously impossible, and we are done. | Assume that <math>m+n\ne 2^k</math>, so then <math>m+n\equiv 0\pmod{p}</math> for some odd prime <math>p</math>. Then <math>m\equiv -n\pmod{p}</math>, so <math>\frac{m}{n}\equiv \frac{n}{m}\equiv -1\pmod{p}</math>. We see that the arithmetic mean is <math>\frac{-1+(-1)}{2}\equiv -1\pmod{p}</math> and the harmonic mean is <math>\frac{2(-1)(-1)}{-1+(-1)}\equiv -1\pmod{p}</math>, so if 1 can be written then <math>1\equiv -1\pmod{p}</math> and <math>2\equiv 0\pmod{p}</math> which is obviously impossible, and we are done. | ||
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This proof is wrong because <math>\frac{m}{n}</math> and <math>\frac{n}{m}</math> are not integers, so you cannot say that they are <math>-1\pmod{p}</math> and work <math>\pmod{p}</math>. | This proof is wrong because <math>\frac{m}{n}</math> and <math>\frac{n}{m}</math> are not integers, so you cannot say that they are <math>-1\pmod{p}</math> and work <math>\pmod{p}</math>. | ||
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{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:34, 18 June 2022
Two rational numbers and
are written on a blackboard, where
and
are relatively prime positive integers. At any point, Evan may pick two of the numbers
and
written on the board and write either their arithmetic mean
or their harmonic mean
on the board as well. Find all pairs
such that Evan can write
on the board in finitely many steps.
Proposed by Yannick Yao
Solution
We claim that all odd work if
is a positive power of 2.
Proof:
We first prove that works. By weighted averages we have that
can be written, so the solution set does indeed work. We will now prove these are the only solutions.
Assume that , so then
for some odd prime
. Then
, so
. We see that the arithmetic mean is
and the harmonic mean is
, so if 1 can be written then
and
which is obviously impossible, and we are done.
-Stormersyle
This proof is wrong because and
are not integers, so you cannot say that they are
and work
.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2019 USAJMO (Problems • Resources) | ||
Preceded by Problem 5 |
Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |