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Revision as of 19:45, 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

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

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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