2022 AMC 10A Problems/Problem 14
Contents
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 by OmegaLearn
~ 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 |
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