2022 AIME I Problems/Problem 6

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

2022 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2022 AIME I Problems/Problem 6

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