Difference between revisions of "2022 AMC 10A Problems/Problem 14"
MRENTHUSIASM (talk | contribs) (→See Also) |
Pi is 3.14 (talk | contribs) |
||
Line 18: | Line 18: | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/V1jOj8ysd_w | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See Also == | == See Also == |
Revision as of 02:57, 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.
- 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.