Difference between revisions of "2022 AMC 10A Problems/Problem 14"
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Revision as of 20:38, 19 September 2023
- The following problem is from both the 2022 AMC 10A #14 and 2022 AMC 12A #10, so both problems redirect to this page.
Contents
[hide]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 1 (Casework)
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.
- If pairs with then can pair with one of After that, each of does not have any restrictions. This case produces ways.
Together, the answer is
~MRENTHUSIASM
Solution 2 (Multiplication Principle)
As said in Solution 1, clearly, the integers from through must be in different pairs.
We know that or can pair with any integer from to , or can pair with any integer from to , and or can pair with any integer from to . Thus, will have choices to pair with, will then have choices to pair with ( cannot pair with the same number as the one pairs with). cannot pair with the numbers and has paired with but can also now pair with , so there are choices. cannot pair with 's, 's, or 's paired numbers, so there will be choices for . can pair with an integer from to that hasn't been paired with already, or it can pair with . will only have one choice left, and must pair with .
So, the answer is
~Scarletsyc
Solution 3 (Generalization)
The integers must each be the larger elements of a distinct pair.
Assign partners in decreasing order for :
Note that must pair with : .
For , the choices are . As decreases by 1, The minuend increases by 2 elements, and the subtrahend increases by 1 element, so the difference increases by 1, yielding .
After assigning a partner to , there are no invalid pairings for yet-unpaired numbers, so there are to choose partners for .
The answer is .
In general, for , the same logic yields answer:
~oinava
Video Solution 1
~Education, the Study of Everything :)
Video Solution by Sohil Rathi
~ pi_is_3.14
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.