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 | ||
