Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
(→‎Solution 2)
Line 16: Line 16:
 
We can solve this problem using casework, with one case for each possible pair of consecutive numbers.
 
We can solve this problem using casework, with one case for each possible pair of consecutive numbers.
  
−
<math>\underline{Case  1: (1,2)}</math>
+
<math>\textbf{Case  1: (1,2)}</math>
  
 
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 <math>8 \choose {0}</math>. We can also pick one number, giving us <math>7 \choose {1}</math> 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 <math>6 \choose {2}</math>. This pattern continues for each amount of numbers, yielding <math>5 \choose {3}</math> for 3 numbers and <math>4 \choose {4}</math> for four numbers. Adding these up, we have <math>8 \choose {0}</math> + <math>7 \choose {1}</math> + <math>6 \choose {2}</math> + <math>5 \choose {3}</math> + <math>4 \choose {4}</math> = <math>\textbf{34}</math>.
 
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 <math>8 \choose {0}</math>. We can also pick one number, giving us <math>7 \choose {1}</math> 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 <math>6 \choose {2}</math>. This pattern continues for each amount of numbers, yielding <math>5 \choose {3}</math> for 3 numbers and <math>4 \choose {4}</math> for four numbers. Adding these up, we have <math>8 \choose {0}</math> + <math>7 \choose {1}</math> + <math>6 \choose {2}</math> + <math>5 \choose {3}</math> + <math>4 \choose {4}</math> = <math>\textbf{34}</math>.

Revision as of 22:18, 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 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 $8 \choose {0}$ + $7 \choose {1}$ + $6 \choose {2}$ + $5 \choose {3}$ + $4 \choose {4}$ = $\textbf{34}$.


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


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.

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)

See also

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
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AIME_I_Problems/Problem_11&oldid=189223"