2023 AIME I Problems/Problem 11

Unofficial problem statement: Let $S$ be the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ . How many subsets of $S$ have exactly one pair of consecutive elements? (Ex: $\{1, 2, 6, 10\}$ )

Solution

Define $f(x)$ to be the number of subsets of $\{1, 2, 3, 4, \ldots x\}$ that have $0$ consecutive element pairs, and $f'(x)$ to be the number of subsets that have $1$ consecutive pair.

Using casework on where the consecutive pair is, it is easy to see that $f'(10) = 2f(7) + 2f(6) + 2f(1)f(5) + 2f(2)f(4) + 2f(3)^2$ .

We see that $f(1) = 2$ , $f(2) = 3$ , and $f(n) = f(n-1) + f(n-2)$ . This is because if the element $1$ is included in our subset, then there are $f(n-2)$ possibilities, and otherwise there are $f(n-1)$ possibilities. Thus, by induction, $f(x)$ is the $n+1$ th Fibonacci number.

This means that $f'(10) = 2(34) + 2(21) + 2(2)(13) + 2(3)(8) + 5^2 = \boxed{235}$ .

~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 $8 \choose {0}$ . We can also pick one number, giving us $7 \choose {1}$ 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 $6 \choose {2}$ . This pattern continues for each amount of numbers, yielding $5 \choose {3}$ for 3 numbers and $4 \choose {4}$ for four numbers. Adding these up, we have $8 \choose {0}$ + $7 \choose {1}$ + $6 \choose {2}$ + $5 \choose {3}$ + $4 \choose {4}$ = 34.

Case 2: (2,3)

Solution 2(Double recursive equations approach)

Denote by $N_1 \left( m \right)$ the number of subsets of a set $S$ that consists of $m$ consecutive integers, such that each subset contains exactly one pair of consecutive integers.

Denote by $N_0 \left( m \right)$ the number of subsets of a set $S$ that consists of $m$ consecutive integers, such that each subset does not contain any consecutive integers.

Denote by $a$ the smallest number in set $S$ .

First, we compute $N_1 \left( m \right)$ .

Consider $m \geq 3$ . We do casework analysis.

Case 1: A subset does not contain $a$ .

The number of subsets that has exactly one pair of consecutive integers is $N_1 \left( m - 1 \right)$ .

Case 2: A subset contains $a$ but does not contain $a + 1$ .

The number of subsets that has exactly one pair of consecutive integers is $N_1 \left( m - 2 \right)$ .

Case 3: A subset contains $a$ and $a + 1$ .

To have exactly one pair of consecutive integers, this subset cannot have $a + 2$ , and cannot have consecutive integers in $\left\{ a+3, a+4, \cdots , a + m - 1 \right\}$ .

Thus, the number of subsets that has exactly one pair of consecutive integers is $N_0 \left( m - 3 \right)$ .

Therefore, for $m \geq 3$ , $N_1 \left( m \right) = N_1 \left( m - 1 \right) + N_1 \left( m - 2 \right) + N_0 \left( m - 3 \right) .$

For $m = 1$ , we have $N_1 \left( 1 \right) = 0$ . For $m = 2$ , we have $N_1 \left( 2 \right) = 1$ .

Second, we compute $N_0 \left( m \right)$ .

Consider $m \geq 2$ . We do casework analysis.

Case 1: A subset does not contain $a$ .

The number of subsets that has no consecutive integers is $N_0 \left( m - 1 \right)$ .

Case 2: A subset contains $a$ .

To avoid having consecutive integers, the subset cannot have $a + 1$ .

Thus, the number of subsets that has no consecutive integers is $N_0 \left( m - 2 \right)$ .

Therefore, for $m \geq 2$ , $N_0 \left( m \right) = N_0 \left( m - 1 \right) + N_0 \left( m - 2 \right) .$

For $m = 0$ , we have $N_0 \left( 0 \right) = 1$ . For $m = 1$ , we have $N_0 \left( 1 \right) = 2$ .

By solving the recursive equations above, we get $N_1 \left( 10 \right) = \boxed{\textbf{(235) }}$ .

~ 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 $1$ way to do this.

Case 1.2: We have 1 other number in our set. Let's say that this number is $x_1$ . We then can say that $x_a = x_1 - 2$ , with the restriction that $x_a \ge 2$ . Then, we can write $x_a \le 8$ (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 $(x_a - 2) \le 6$ , where $x_a \ge 0$ . We then can use stars and bars: $7 \choose {1}$ $= 7$ .

Case 1.3: We have 2 numbers in our set. This is similar, so I won't go through the same process. $x_a = x_1 - 2, x_b = x_2 - x_1, x_{a, b} \ge 2$ . $x_a + x_b \le 8, (x_a - 2) + (x_b - 2) \le 4$ . We can use stars and bars: $6 \choose 2$ $= 15$ . Case 1.4: We have 3 numbers in our set. $x_a + x_b + x_c \le 8, (x_a - 2) + (x_b - 2) + (x_c - 2) \le 2$ . Stars and bars: $5 \choose 3$ $= 10$ .

Case 1.4: We have 4 numbers in our set. $x_a + x_b + x_c + x_d \le 8, (x_a - 2) + (x_b - 2) + (x_c - 2) + (x_d - 2) \le 0$ . Stars and bars: $4 \choose 4$ $= 1$ .

Adding all subcases, we get $1 + 7 + 15 + 10 + 1 = 34$ .

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 $34$ cases.

Case 3: Our consecutive pair is (2,3). The value is $21$ 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 $21$ . 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

https://youtu.be/7xqeCQiPrew

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

2023 AIME I (Problems • Answer Key • Resources)
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