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 (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)

Video Solution

https://youtu.be/7xqeCQiPrew

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2023 AIME I Problems/Problem 11

Solution

Solution (Double recursive equations approach)

Video Solution