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(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
Solution 3
Consider the number of binary strings that are length that satisfy certain properties. How do you make binary strings such that
's don't touch each other? You either append
to binary string that is
less length or
to binary string that is
less length. You end up with Fibonacci sequence. The
th fibonacci is
and
so
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)