2023 AIME I Problems/Problem 11
Unofficial problem statement: Let be the set . How many subsets of have exactly one pair of consecutive elements? (Ex: )
Contents
Solution
Define to be the number of subsets of that have consecutive element pairs, and to be the number of subsets that have consecutive pair.
Using casework on where the consecutive pair is, it is easy to see that .
We see that , , and . This is because if the element is included in our subset, then there are possibilities, and otherwise there are possibilities. Thus, by induction, is the th Fibonacci number.
This means that .
~mathboy100
Solution 2
We can solve this problem using casework, with one case for each possible pair of consecutive numbers.
Case 1: (1,2)
If we have (1,2) as our pair, we are left with the numbers from 3-10 as elements that can be added to our subset. So, we must compute how many ways we can pick these numbers so that the set has no consecutive numbers other than (1,2). Our first option is to pick no more numbers, giving us . We can also pick one number, giving us because 3 cannot be picked.
Solution 2(Double recursive equations approach)
Denote by the number of subsets of a set that consists of consecutive integers, such that each subset contains exactly one pair of consecutive integers.
Denote by the number of subsets of a set that consists of consecutive integers, such that each subset does not contain any consecutive integers.
Denote by the smallest number in set .
First, we compute .
Consider . We do casework analysis.
Case 1: A subset does not contain .
The number of subsets that has exactly one pair of consecutive integers is .
Case 2: A subset contains but does not contain .
The number of subsets that has exactly one pair of consecutive integers is .
Case 3: A subset contains and .
To have exactly one pair of consecutive integers, this subset cannot have , and cannot have consecutive integers in .
Thus, the number of subsets that has exactly one pair of consecutive integers is .
Therefore, for ,
For , we have . For , we have .
Second, we compute .
Consider . We do casework analysis.
Case 1: A subset does not contain .
The number of subsets that has no consecutive integers is .
Case 2: A subset contains .
To avoid having consecutive integers, the subset cannot have .
Thus, the number of subsets that has no consecutive integers is .
Therefore, for ,
For , we have . For , we have .
By solving the recursive equations above, we get .
~ Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3
Let's consider all cases:
Case 1: Our consecutive pair is (1,2).
Then, let's make subcases:
Case 1.1: We have 0 other numbers in our set. There is way to do this.
Case 1.2: We have 1 other number in our set. Let's say that this number is . We then can say that , with the restriction that . Then, we can write (since we have a total of 8 spots for remaining numbers: 3, 4, 5, 6, 7, 8, 9, 10). To remove the restriction on x_a, we can write , where . We then can use stars and bars: .
Case 1.3: We have 2 numbers in our set. This is similar, so I won't go through the same process. . . We can use stars and bars: . Case 1.4: We have 3 numbers in our set. . Stars and bars: .
Case 1.4: We have 4 numbers in our set. . Stars and bars: .
Adding all subcases, we get .
Case 2: Our consecutive pair is (8,9).
We use symmetry to find that there is exactly 1 case for (8,9) that matches with (1,2). This also has cases.
Case 3: Our consecutive pair is (2,3). The value is This is left as an exercise to the reader (but I can write it if you'd like).
Case 4: Our consecutive pair is (7,8). The value is . This is left as an exercise to the reader (but I can write it if you'd like).
- I will finish later.
hia2020
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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