2023 AIME I Problems/Problem 11
Contents
- 1 Problem
- 2 Solution 1 (Minimal Casework)
- 3 Solution 2
- 4 Solution 3 (Double Recursive Equations)
- 5 Solution 5 (Similar to Solution 3)
- 6 Solution 6 (Stars and Bars)
- 7 Solution 7 (Fibonacci)
- 8 Solution 8 (Polyominoes)
- 9 Solution 9 (Dynamic Programming)
- 10 Video Solution
- 11 Video Solution (Mathematical Dexterity)
- 12 See also
Problem
Find the number of subsets of that contain exactly one pair of consecutive integers. Examples of such subsets are and
Solution 1 (Minimal Casework)
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 element pair is, there is a unique consecutive element pair that satisfies the conditions. 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 for the rest of the elements (because cannot be used), 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.
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. Another choice is to pick two numbers and in order to make sure they are not consecutive we must fix one number in between them, giving us . This pattern continues for each amount of numbers, yielding for 3 numbers and for four numbers. Adding these up, we have + + + + = .
If we have (2,3) as our pair, everything works the same as with (1,2), because 1 is still unusable as it is consecutive with 2. The only difference is we now have only 4-10 to work with. Using the same pattern as before, we have + + + = .
This case remains pretty much the same except we now have an option of whether or not to include 1. If we want to represent this like we have with our other choices, we would say for choosing no numbers and for choosing 1, leaving us with + = 2 choices (either including the number 1 in our subset or not including it). As far as the numbers from 5-10, our pattern from previous cases still holds. We have + + + = 13. With 2 choices on one side and 13 choices on the other side, we have = combinations in all.
Following the patterns we have already created in our previous cases, for the numbers 1-3 we have + = 3 choices (1, 2, or neither) and for the numbers 6-10 we have + + = 8 choices. With 3 choices on one side and 8 choices on the other side, we have = combinations in all.
Again following the patterns we have already created in our previous cases, for the numbers 1-4 we have + + = 5 choices and for the numbers 5-10 we have the same + + = 5 choices. = combinations in all.
By symmetry, the case with (6,7) will act the same as case 4 with (4,5). This goes the same for (7,8) and case 3, (8.9) and case 2, and (9,10) and case 1.
Now, we simply add up all of the possibilities for each case to get our final answer. 34 + 21 + 26 + 24 + 25 + 24 + 26 + 21 + 34 =
-Algebraik
Solution 3 (Double Recursive Equations)
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 5 (Similar to Solution 3)
Let be the number of subsets of the set such that there exists exactly 1 pair of consecutive elements. Let be the number of subsets of the set such that there doesn't exist any pair of consecutive elements. First, lets see how we can construct For each subset counted in either: 1. 2. , or 3. and The first case counts subsets (as cannot be included and the rest cannot have any consecutive elements), The second counts and the third counts Thus, Next, Lets try to construct For each subset counted in either: 1. or 2. The first case counts subsets and the second counts Thus, Since and we have that so (The is the th Fibonacci number). From here, we can construct a table of the values of until By listing out possibilities, we can solve for our first 3 values.
Our answer is ~AtharvNaphade
Solution 6 (Stars and Bars)
Note: This is a very common stars and bars application.
Casework on number of terms, let the number of terms be . We can come up with a generalized formula for the number of subsets with n terms.
Let be the differences between the n terms. For example, in the set {2, 3, 6},
Let the range of the set be k for now, . We select one pair of terms to be consecutive by selecting one of the (n-1) terms to be 1. WLOG, let . .
To ensure the other are greater than 1 such that no two other terms are consecutive or the same, let . where are positive integers.
Finally, we add in , the distance between 0 and the first term of the set, and , the distance between the last term and 11. This way, the "distance" from 0 to 11 is "bridged" by , k, and . There are n positive terms, by Balls and Urns, there are ways of doing this. However, recall that there were ways, and we had used a WLOG to choose which two digits are consecutive. The final formula for the number of valid n-element subsets is hence for .
Case 1: Two terms , so
Case 2: Three terms , so
Case 3: Four terms , so
Case 4: Five terms , so
Case 5: Six terms , so
We can check by the Pigeonhole principle that there cannot be more than six terms, so the answer is .
~Mathandski
Solution 7 (Fibonacci)
Note that there are subsets of a set of consecutive integers that contains no two consecutive integers. (This can be proven by induction.)
Now, notice that if we take and as the consecutive integers in our subset, we need to make a subset of the remaining integers such that it doesn't contain any two consecutive integers. Clearly, and cannot be chosen, and since and are sufficiently far apart, it is obvious we do not need to be concerned that an element of the set is consecutive with any element of the set
Thus, we can count the number of ways to choose a subset from the first set without any two elements being consecutive and multiply this by the number of ways to choose a subset from the second set without any two elements being consecutive. From above, and noting that the first set has consecutive integer elements and the second set has consecutive integer elements, we know that this is
Summing this over for all yields
~lpieleanu
Solution 8 (Polyominoes)
The problem is the same as laying out a line of polynomoes to cover spots : 1 triomino (), dominoes (), and monominoes (). The spots cover the members of the subset. The total number spots is 11, because one spot always covers the 0, and the other spots cover 1 through 10.
There are 5 ways to choose polyomino sets, and many ways to order each set:
Polyominoes Orderings
The sum is .
~BraveCobra22aops
Solution 9 (Dynamic Programming)
Let be the number of subsets of a set consecutive integers such that the maximum value in the set is and there exists exactly one pair of consecutive integers. Define similarly, but without any pair of consecutive integers. The base cases are , , and .
The transitions are:
Note that is the Fibonacci numbers.
Summing over yields
~Mathenthus
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=q1ffZP63q3I
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
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Followed by Problem 12 | |
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