Difference between revisions of "2019 USAJMO Problems/Problem 5"
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Revision as of 22:51, 19 April 2019
Let be a nonnegative integer. Determine the number of ways that one can choose
sets
, for integers
with
, such that:
1. for all , the set
has
elements; and
2. whenever
and
.
Proposed by Ricky Liu
Solution
Note that there are ways to choose
, because there are
ways to choose which number
is,
ways to choose which number to append to make
,
ways to choose which number to append to make
... After that, note that
contains the
in
and 1 other element chosen from the 2 elements in
not in
so there are 2 ways for
. By the same logic there are 2 ways for
as well so
total ways for all
, so doing the same thing
more times yields a final answer of
.
-Stormersyle
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 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |