Difference between revisions of "2022 AMC 10A Problems/Problem 14"
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{{AMC10 box|year=2022|ab=A|num-b=13|num-a=15}} | {{AMC10 box|year=2022|ab=A|num-b=13|num-a=15}} | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 01:33, 12 November 2022
Problem
How many ways are there to split the integers 
through 
into 
pairs such that in each pair, the greater number is at least 
times the lesser number?
Solution
Clearly, the integers from 
through must be in different pairs, and must pair with 
Note that 
can pair with either 
or 
From here, we consider casework:
- If
pairs with 
then 
can pair with one of 
After that, each of 
does not have any restrictions. This case produces 
ways.
then 
can pair with one of 
After that, each of 
does not have any restrictions. This case produces 
ways.- If pairs with
then can pair with one of 
After that, each of does not have any restrictions. This case produces ways.
then 
After that, each of Together, the answer is 
~MRENTHUSIASM
See Also
| 2022 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2022 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
