2023 AMC 12A Problems/Problem 24
Contents
Problem
Let be the number of sequences , , , such that is a positive integer less than or equal to , each is a subset of , and is a subset of for each between and , inclusive. For example, , , , , is one such sequence, with .What is the remainder when is divided by ?
Solution 1
Consider any sequence with terms. Every 10 number has such choices: never appear, appear the first time in the first spot, appear the first time in the second spot… and appear the first time in the th spot, which means every number has choices to show up in the sequence. Consequently, for each sequence with length , there are possible ways.
Thus, the desired value is
~bluesoul
Solution 2
Let be the number of sequences , , , such that each is a subset of , and is a subset of for , , . Then and .
If and , we need to get a recursive formula for : If , then has possibilities, and the subsequence has possibilities. Hence By applying this formula and only considering modulo , we get , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , .
Lastly, we get . ~Quantum-Phantom
Solution 3 (Cheese, but this time it actually works)
Seeing that all the answers are different modulus 5, and that 10 is divisible by 5, we cheese this problem.
Let be one sequence satisfying the constraints of the problem. Let be the sequence of nonnegative integers such that has elements for all , and . Note that we can generate the number of valid sequences of by first generating all sequences of such that for all , then choosing the elements from that we keep in , given the sequence of as the restraint for the number of elements. For each sequence , there are corresponding sequences for . Now, consider two cases - either all terms in are either 0, 5, or 10, or there is at least one term in that is neither 0, 5 nor 10. In the second case, consider the last term in that is not 10 or 5, say . However, that implies , and so the number of corresponding sequences of is something or something, which is always a multiple of . Therefore, we only need to consider sequences of where each term is or . If all terms in are 0 or 10, then for each there are sequences of (since there are places to turn from to ), for a total of (mod 5). If there exists at least one term , then we use stars and bars to count the number of sequences of , and each sequence of corresponds to sequences of . For each , we must have at least one term of . After that, there are stars and bars (separating to and to ), so that is sequences of . So the sum is (mod 5). Therefore, the answer is 0 mod 5, and it must be
Solution
We observe that in each sequence, if element , then for all . Therefore, to determine a sequence with a fixed length , we only need to determine the first set that each element in is inserted into, or an element is never inserted into any subset.
We have
Recalling or noticing that , then, Modulo 10, we have
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Slightly easier, observe that , so, working , we have
~oinava
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2023 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.