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

(Created page with "Unofficial problem statement: Let <math>S</math> be the set <math>{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}</math>. How many subsets of <math>S</math> have exactly one pair of consecuti...")
 
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Unofficial problem statement: Let <math>S</math> be the set <math>{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}</math>. How many subsets of <math>S</math> have exactly one pair of consecutive elements? (Ex: <math>(1, 2, 6, 10)</math>)
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Unofficial problem statement: Let <math>S</math> be the set <math>\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}</math>. How many subsets of <math>S</math> have exactly one pair of consecutive elements? (Ex: <math>(1, 2, 6, 10)</math>)
  
 
==Solution==
 
==Solution==
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Define <math>f(x)</math> to be the number of subsets of <math>{1, 2, 3, 4, 5, 6, 7}</math> that have <math>0</math> consecutive element pairs, and <math>f'(x)</math> to be the number of subsets that have <math>1</math> consecutive pair.
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Define <math>f(x)</math> to be the number of subsets of <math>\{1, 2, 3, 4, \ldots x\}</math> that have <math>0</math> consecutive element pairs, and <math>f'(x)</math> to be the number of subsets that have <math>1</math> consecutive pair.
  
 
Using casework on where the consecutive pair is, it is easy to see that <math>f'(10) = 2f(7) + 2f(6) + 2f(1)f(5) + 2f(2)f(4) + 2f(3)^2</math>.
 
Using casework on where the consecutive pair is, it is easy to see that <math>f'(10) = 2f(7) + 2f(6) + 2f(1)f(5) + 2f(2)f(4) + 2f(3)^2</math>.

Revision as of 12:48, 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

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