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Difference between revisions of "2022 AIME I Problems/Problem 6"

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| S | - \left( | A | + | B | + | C | \right) = 276 - \left( 30 + 12 + 6 \right) = \boxed{228}.
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| S | - \left( | A | + | B | + | C | \right)
 
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& = 276 - \left( 30 + 12 + 6 \right) \\
 
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Revision as of 23:43, 17 February 2022

Problem

Find the number of ordered pairs of integers $(a, b)$ such that the sequence\[3, 4, 5, a, b, 30, 40, 50\]is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.

Solution 1

Since $3,4,5,a$ and $3,4,5,b$ cannot be an arithmetic progression, $a$ or $b$ can never be $6$. Since $b, 30, 40, 50$ and $a, 30, 40, 50$ cannot be an arithmetic progression, $b$ can never be $20$. Since $a < b$, there are ${24 - 2 \choose 2} = 231$ ways to choose $a$ and $b$ with these two restrictions in mind.

However, there are still specific invalid cases counted in these $231$ pairs $(a,b)$. Since \[3,5,a,b\] cannot form an arithmetic progression, $\underline{(a,b) \neq (7,9)}$. \[a,b,30,50\] cannot be an arithmetic progression, so $(a,b) \neq (-10,10)$; however, since this pair was not counted in our $231$, we do not need to subtract it off. \[3,a,b,30\] cannot form an arithmetic progression, so $\underline{(a,b) \neq (12,21)}$. \[4, a, b, 40\] cannot form an arithmetic progression, so $\underline{(a,b) \neq (16,28)}$. \[5, a,b, 50\] cannot form an arithmetic progression, $(a,b) \neq 20, 35$; however, since this pair was not counted in our $231$ (since we disallowed $a$ or $b$ to be $20$), we do not to subtract it off.

Also, the sequences $(3,a,b,40)$, $(3,a,b,50)$, $(4,a,b,30)$, $(4,a,b,50)$, $(5,a,b,30)$ and $(5,a,b,40)$ will never be arithmetic, since that would require $a$ and $b$ to be non-integers.

So, we need to subtract off $3$ progressions from the $231$ we counted, to get our final answer of $\boxed{228}$.

~ ihatemath123

Solution 2

Denote $S = \left\{ (a, b) : 6 \leq a < b \leq 29 \right\}$.

Denote by $A$ a subset of $S$, such that there exists an arithmetic sequence that has 4 terms and includes $a$ but not $b$.

Denote by $B$ a subset of $S$, such that there exists an arithmetic sequence that has 4 terms and includes $b$ but not $a$.

Hence, $C$ is a subset of $S$, such that there exists an arithmetic sequence that has 4 terms and includes both $a$ and $b$.

Hence, this problem asks us to compute \[ | S | - \left( | A | + | B | + | C | \right) . \]

First, we compute $| S |$.

We have $| S | = \binom{29 - 6 + 1}{2} = \binom{24}{2} = 276$.

Second, we compute $| A |$.

$\textbf{Case 1}$: $a = 6$.

We have $b = 8 , \cdots , 19, 21, 22, \cdots, 29$. Thus, the number of solutions is 21.

$\textbf{Case 2}$: $a = 20$.

We have $b = 21, 22, \cdots , 29$. Thus, the number of solutions is 9.

Thus, $| A | = 21 + 9 = 30$.

Third, we compute $| B |$.

In $B$, we have $b = 6, 20$. However, because $6 \leq a < b$, we have $b \geq 7$. Thus, $b = 20$.

This implies $a = 7, 8, 9, 11, 12, \cdots , 19$. Thus, $| B | = 12$.

Fourth, we compute $| C |$.

$\textbf{Case 1}$: In the arithmetic sequence, the two numbers beyond $a$ and $b$ are on the same side of $a$ and $b$.

Hence, $(a, b) = (6 , 7), (7, 9) , (10, 20)$. Therefore, the number solutions in this case is 3.

$\textbf{Case 2}$: In the arithmetic sequence, the two numbers beyond $a$ and $b$ are on the opposite sides of $a$ and $b$.

$\textbf{Case 2.1}$: The arithmetic sequence is $3, a, b, 30$.

Hence, $(a, b) = (12, 21)$.

$\textbf{Case 2.2}$: The arithmetic sequence is $4, a, b, 40$.

Hence, $(a, b) = (16, 28)$.

$\textbf{Case 2.3}$: The arithmetic sequence is $5, a, b, 50$.

Hence, $(a, b) = (20, 35)$.

Putting two cases together, $| C | = 6$.

Therefore, \[| S | - \left( | A | + | B | + | C | \right) = 276 - \left( 30 + 12 + 6 \right) = \boxed{228}.\]

~Steven Chen (www.professorchenedu.com)

See Also

2022 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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