Difference between revisions of "2023 AIME I Problems/Problem 11"

Revision as of 21:35, 8 February 2023

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 1

We can solve this problem using casework, with one case for each possible pair of consecutive numbers.

Case 1:

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)

@@ Line 14: / Line 14: @@
 ==Solution 1==
-test
+We can solve this problem using casework, with one case for each possible pair of consecutive numbers.
+Case 1:
 ==Solution 2(Double recursive equations approach)==

2023 AIME I (Problems • Answer Key • Resources)
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