Difference between revisions of "2022 AMC 12A Problems/Problem 24"
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− | must also belong in <math>S</math>. | + | must also belong in <math>S</math>. However, one must consider the edge case all 5 elements are the same (only <math>\{0,0,0,0,0\}</math>), in which case all sequences listed are equivalent. Then <math>\lvert S \rvert \equiv 1 \pmod 5</math>, which by inspection of the answer choices yields <math>\boxed{\textbf{(E) 1296}}</math>. |
~Tau | ~Tau |
Revision as of 12:54, 12 November 2022
Contents
[hide]Problem
How many strings of length formed from the digits , , , , are there such that for each , at least of the digits are less than ? (For example, satisfies this condition because it contains at least digit less than , at least digits less than , at least digits less than , and at least digits less than . The string does not satisfy the condition because it does not contain at least digits less than .)
Solution
Denote by the number of -digit strings formed by using numbers , where for each , at least of the digits are less than .
We have the following recursive equation: and the boundary condition for any .
By solving this recursive equation, for and , we get
For and , we get
For and , we get
For and , we get
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (Cheese)
Let the set of all valid sequences be . Notice that for any sequence in , the sequences must also belong in . However, one must consider the edge case all 5 elements are the same (only ), in which case all sequences listed are equivalent. Then , which by inspection of the answer choices yields .
~Tau
Video Solution By ThePuzzlr
~ MathIsChess
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution By OmegaLearn using Complementary Counting
~ pi_is_3.14
See Also
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.