Art of Problem Solving
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

2023 AIME I Problems/Problem 11: Difference between revisions

← Older editNewer edit →
Mathboy100 (talk | contribs)
editor
381 edits
No edit summary
No edit summary
Line 11: Line 11:


~mathboy100
~mathboy100
==Solution (Double recursive equations approach)==
Denote by <math>N_1 \left( m \right)</math> the number of subsets of a set <math>S</math> that consists of <math>m</math> consecutive integers, such that each subset contains exactly one pair of consecutive integers.
Denote by <math>N_0 \left( m \right)</math> the number of subsets of a set <math>S</math> that consists of <math>m</math> consecutive integers, such that each subset does not contain any consecutive integers.
Denote by <math>a</math> the smallest number in set <math>S</math>.
First, we compute <math>N_1 \left( m \right)</math>.
Consider <math>m \geq 3</math>.
We do casework analysis.
Case 1: A subset does not contain <math>a</math>.
The number of subsets that has exactly one pair of consecutive integers is <math>N_1 \left( m - 1 \right)</math>.
Case 2: A subset contains <math>a</math> but does not contain <math>a + 1</math>.
The number of subsets that has exactly one pair of consecutive integers is <math>N_1 \left( m - 2 \right)</math>.
Case 3: A subset contains <math>a</math> and <math>a + 1</math>.
To have exactly one pair of consecutive integers, this subset cannot have <math>a + 2</math>, and cannot have consecutive integers in <math>\left\{ a+3, a+4, \cdots , a + m - 1 \right\}</math>.
Thus, the number of subsets that has exactly one pair of consecutive integers is <math>N_0 \left( m - 3 \right)</math>.
Therefore, for <math>m \geq 3</math>,
<cmath>
N_1 \left( m \right) = N_1 \left( m - 1 \right) + N_1 \left( m - 2 \right) + N_0 \left( m - 3 \right) .
</cmath>
For <math>m = 1</math>, we have <math>N_1 \left( 1 \right) = 0</math>.
For <math>m = 2</math>, we have <math>N_1 \left( 2 \right) = 1</math>.
Second, we compute <math>N_0 \left( m \right)</math>.
Consider <math>m \geq 2</math>.
We do casework analysis.
Case 1: A subset does not contain <math>a</math>.
The number of subsets that has no consecutive integers is <math>N_0 \left( m - 1 \right)</math>.
Case 2: A subset contains <math>a</math>.
To avoid having consecutive integers, the subset cannot have <math>a + 1</math>.
Thus, the number of subsets that has no consecutive integers is <math>N_0 \left( m - 2 \right)</math>.
Therefore, for <math>m \geq 2</math>,
<cmath>
N_0 \left( m \right) = N_0 \left( m - 1 \right) + N_0 \left( m - 2 \right)  .
</cmath>
For <math>m = 0</math>, we have <math>N_0 \left( 0 \right) = 1</math>.
For <math>m = 1</math>, we have <math>N_0 \left( 1 \right) = 2</math>.
By solving the recursive equations above, we get <math>N_1 \left( 10 \right) = \boxed{\textbf{(235) }}</math>.
~ Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
==Video Solution==
https://youtu.be/7xqeCQiPrew

Revision as of 12:25, 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 (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

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2023_AIME_I_Problems/Problem_11&oldid=188861"