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

Revision as of 21:18, 13 February 2023

Problem

Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$

Solution 1 (Minimal Casework)

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 element pair is, it is easy to see that $f'(10) = 2f(7) + 2f(6) + 2f(1)f(5) + 2f(2)f(4) + f(3)^2.$

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 for the rest of the elements (because $2$ cannot be used), 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 ~lpieleanu (edits)

Solution 2

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

$\textbf{Case 1: (1,2)}$

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 $8 \choose {0}$ + $7 \choose {1}$ + $6 \choose {2}$ + $5 \choose {3}$ + $4 \choose {4}$ = $\textbf{34}$ .

$\textbf{Case 2: (2,3)}$

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 $7 \choose {0}$ + $6 \choose {1}$ + $5 \choose {2}$ + $4 \choose {3}$ = $\textbf{21}$ .

$\textbf{Case 3: (3,4)}$

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 $2 \choose {0}$ for choosing no numbers and $1 \choose {1}$ for choosing 1, leaving us with $2 \choose {0}$ + $1 \choose {1}$ = 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 $6 \choose {0}$ + $5 \choose {1}$ + $4 \choose {2}$ + $3 \choose {3}$ = 13. With 2 choices on one side and 13 choices on the other side, we have $2\cdot13$ = $\textbf{26}$ combinations in all.

$\textbf{Case 4: (4,5)}$

Following the patterns we have already created in our previous cases, for the numbers 1-3 we have $3 \choose {0}$ + $2 \choose {1}$ = 3 choices (1, 2, or neither) and for the numbers 6-10 we have $5 \choose {0}$ + $4 \choose {1}$ + $3 \choose {2}$ = 8 choices. With 3 choices on one side and 8 choices on the other side, we have $3\cdot8$ = $\textbf{24}$ combinations in all.

$\textbf{Case 5: (5,6)}$

Again following the patterns we have already created in our previous cases, for the numbers 1-4 we have $4 \choose {0}$ + $3 \choose {1}$ + $2 \choose {2}$ = 5 choices and for the numbers 5-10 we have the same $4 \choose {0}$ + $3 \choose {1}$ + $2 \choose {2}$ = 5 choices. $5\cdot5$ = $\textbf{25}$ combinations in all.

$\textbf{Rest of the cases}$

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 = $\boxed{\textbf{(235)}}$

-Algebraik

Solution 3 (Double Recursive Equations)

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 4

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

Solution 5 (Similar to Solution 3)

Let $a_n$ be the number of subsets of the set $\{1,2,3,\ldots,n\}$ such that there exists exactly 1 pair of consecutive elements. Let $b_n$ be the number of subsets of the set $\{1, 2, 3\ldots, n\}$ such that there doesn't exist any pair of consecutive elements. First, lets see how we can construct $a_n.$ For each subset $S$ counted in $a_n,$ either: 1. $\{n-1, n\}\subseteq S,$ 2. $n\not\in S$ , or 3. $n-1 \not\in S$ and $n\in S.$ The first case counts $b_{n-3}$ subsets (as $n-1$ cannot be included and the rest cannot have any consecutive elements), The second counts $a_{n-1},$ and the third counts $a_{n-2}.$ Thus, $a_n = a_{n-1} + a_{n-2} + b_{n-3}.$ Next, Lets try to construct $b_n.$ For each subset $T$ counted in $b_n,$ either: 1. $n \not\in T,$ or 2. $n \in T.$ The first case counts $b_{n-1}$ subsets and the second counts $b_{n-2}.$ Thus, $b_n = b_{n-1} + b_{n-2}.$ Since $b_1 = 2$ and $b_2 = 3,$ we have that $b_n = F_{n+1},$ so $a_n = a_{n-1} + a_{n-2} + F_{n-2}.$ (The $F_i$ is the $i$ th Fibonacci number). From here, we can construct a table of the values of $a_n$ until $n = 10.$ By listing out possibilities, we can solve for our first 3 values.

$\begin{array}{r|l} n & a_n \\ \hline 1 & 0 \\ 2 & 1\\ 3 & 2\\ 4 & 2 + 1 + F_2 = 5\\5 & 5 + 2 + F_3 = 10\\6 & 10 + 5 + F_4 = 20 \\ 7 & 20 + 10 + F_5 = 38\\ 8 & 38 + 20 + F_6 = 71 \\ 9& 71 + 38 + F_7 = 130\\ 10 & 130 + 71 + F_8 = 235 \end{array}$ Our answer is $a_{10} = \boxed{235}.$ ~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 $n$ . We can come up with a generalized formula for the number of subsets with n terms.

Let $d_1, d_2, ..., d_{n-1}$ be the differences between the n terms. For example, in the set {2, 3, 6}, $d_1 = 1; d_2 = 3; d_3 = 5$

Let the range of the set be k for now, $d_1 + ... + d_{n-1} = k$ . We select one pair of terms to be consecutive by selecting one of the (n-1) terms to be 1. WLOG, let $d_1 = 1$ . $1 + d_2 + ... + d_{n-1} = k$ .

To ensure the other $d_2, d_3, ..., d_{n-1}$ are greater than 1 such that no two other terms are consecutive or the same, let $D_2 = d_2 + 1; D_3 = d_3 + 1; ...$ . $(n-2) + 1 + D_2 + ... + D_{n-1} = k$ where $D_2, D_3, ..., D_{n-1}$ are positive integers.

Finally, we add in $D_0$ , the distance between 0 and the first term of the set, and $D_n$ , the distance between the last term and 11. This way, the "distance" from 0 to 11 is "bridged" by $D_0$ , k, and $D_N$ . $D_0 + k + D_n = D_0 + ((n-1) + D_2 + ... + D_{n-1}) + D_N = 11$ $D_0 + D_2 + D_3 + ... + D_n = 12 - n$ There are n positive terms, by Balls and Urns, there are ${(12-n)-1 \choose (n)-1} = {11-n \choose n-1}$ ways of doing this. However, recall that there were $(n-1)$ 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 $(n-1){11-n \choose n-1}$ for $n > 2$ .

Case 1: Two terms $n = 2$ , so $(2-1){11-2 \choose 2-1} = 1{9 \choose 1} = 9$

Case 2: Three terms $n = 3$ , so $(3-1){11-3 \choose 3-1} = 2{8 \choose 2} = 56$

Case 3: Four terms $n = 4$ , so $(4-1){11-4 \choose 4-1} = 3{7 \choose 3} = 105$

Case 4: Five terms $n = 5$ , so $(5-1){11-5 \choose 5-1} = 4{6 \choose 4} = 60$

Case 5: Six terms $n = 6$ , so $(6-1){11-6 \choose 6-1} = 5{5 \choose 5} = 5$

We can check by the Pigeonhole principle that there cannot be more than six terms, so the answer is $9+56+105+60+5=\boxed{235}$ .

~Mathandski

Solution 7 (Fibonacci, Quick)

Note that there are $F_{n+2}$ subsets of a set of $n$ consecutive integers that contains no two consecutive integers. (This can be proven by induction.)

Now, notice that if we take $i$ and $i+1$ 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, $i-1$ and $i+2$ cannot be chosen, and since $i-2$ and $i+3$ are sufficiently far apart, it is obvious we do not need to be concerned that an element of the set $\{1, 2, ... , i-2 \}$ is consecutive with any element of the set $\{ i+3, i+4, ... , 10 \}$

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 $i$ consecutive integer elements and the second set has $10-i$ consecutive integer elements, we know that this is $F_i F_{10-i}.$

Summing this over for all $1 \leq i \leq 9$ yields $\sum_{i=1}^9 F_i F_{10-i} = F_1F_9 + F_2F_8 + F_3F_7 + F_4F_6 + F_5 F_5 + F_6 F_4 + F_7 F_3 + F_8 F_2 + F_9 F_1 = 2(F_1F_9 + F_2F_8 + F_3F_7+F_4F_6) + F_5^2 = 2(1 \cdot 34 + 1 \cdot 21 + 2 \cdot 13 + 3 \cdot 8) + 5^2 = 2 \cdot 105+25 = \boxed{235}.$

~lpieleanu

Solution (Polyominoes)

The problem is the same as laying out a line of polynomoes to cover spots $0,1,...10$ : 1 triomino ( $RGG$ ), $n$ dominoes ( $RG$ ), and $8-2n$ monominoes ( $R$ ). The $G$ spots cover the members of the subset. The total number spots is 11, because one $R$ 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:

$R + RG + RGG =$ Polyominoes $\rightarrow$ Orderings

$0 + 4 + 1 = 5 \rightarrow 5! / 0!4!1! = ~~~5$

$2 + 3 + 1 = 6 \rightarrow 6! / 2!3!1! = ~60$

$4 + 2 + 1 = 7 \rightarrow 7! / 4!2!1! = 105$

$6 + 1 + 1 = 8 \rightarrow 8! / 6!1!1! = ~56$

$8 + 0 + 1 = 9 \rightarrow 9! / 8!0!1! = ~~~9$

The sum is $\boxed{235}$ .

~BraveCobra22aops

Video Solution

https://youtu.be/7xqeCQiPrew

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

Video Solution (Mathematical Dexterity)

https://www.youtube.com/watch?v=q1ffZP63q3I

@@ Line 229: / Line 229: @@
 <math>R + RG + RGG = </math> Polyominoes <math>\rightarrow</math> Orderings
-<math>0 + 4 + 1 = 5 \rightarrow 5! / 0!4!1! = 5 </math>
+<math>0 + 4 + 1 = 5 \rightarrow 5! / 0!4!1! = ~~~5 </math>
-<math>2 + 3 + 1 = 6 \rightarrow 6! / 2!3!1! = 60 </math>
+<math>2 + 3 + 1 = 6 \rightarrow 6! / 2!3!1! = ~60 </math>
 <math>4 + 2 + 1 = 7 \rightarrow 7! / 4!2!1! = 105 </math>
-<math>6 + 1 + 1 = 8 \rightarrow 8! / 6!1!1! = 56 </math>
+<math>6 + 1 + 1 = 8 \rightarrow 8! / 6!1!1! = ~56 </math>
-<math>8 + 0 + 1 = 9 \rightarrow 9! / 8!0!1! = 9 </math>
+<math>8 + 0 + 1 = 9 \rightarrow 9! / 8!0!1! = ~~~9 </math>
 The sum is <math>\boxed{235}</math>.

2023 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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