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Difference between revisions of "2022 AIME I Problems/Problem 6"
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==solution 1==
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== Problem ==
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Find the number of ordered pairs of integers <math>(a, b)</math> such that the sequence<cmath>3, 4, 5, a, b, 30, 40, 50</cmath>is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.
  
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divide cases into <math>7\leq a<20; 21\leq a\leq28</math>.
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== Solution ==
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There are three cases that arithmetic sequence forms: <math>3,12,21,30;4,16,28,40;3,5,7,9</math>. So the answer is <math>22+...+10+1+2+...+8-13-3=228</math>
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Since <math>3,4,5,a</math> and <math>3,4,5,b</math> cannot be an arithmetic progression, <math>a</math> or <math>b</math> can never be <math>6</math>. Since <math>b, 30, 40, 50</math> and <math>a, 30, 40, 50</math> cannot be an arithmetic progression, <math>b</math> can never be <math>20</math>. Since <math>a < b</math>, there are <math>{24 - 2 \choose 2} = 231</math> ways to choose <math>a</math> and <math>b</math> with these two restrictions in mind.
  
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~bluesoul
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However, there are still specific invalid cases counted in these <math>231</math> pairs <math>(a,b)</math>. Since
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<cmath>3,5,a,b</cmath>
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cannot form an arithmetic progression, <math>\underline{(a,b) \neq (7,9)}</math>.
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<cmath>a,b,30,50</cmath>
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cannot be an arithmetic progression, so <math>(a,b) \neq (-10,10)</math>; however, since this pair was not counted in our <math>231</math>, we do not need to subtract it off.
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<cmath>3,a,b,30</cmath>
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cannot form an arithmetic progression, so <math>\underline{(a,b) \neq (12,21)}</math>.
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<cmath>4, a, b, 40</cmath>
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cannot form an arithmetic progression, so <math>\underline{(a,b) \neq (16,28)}</math>.
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<cmath>5, a,b, 50</cmath>
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cannot form an arithmetic progression, <math>(a,b) \neq 20, 35</math>; however, since this pair was not counted in our <math>231</math> (since we disallowed <math>a</math> or <math>b</math> to be <math>20</math>), we do not to subtract it off.
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Also, the sequences <math>(3,a,b,40)</math>, <math>(3,a,b,50)</math>, <math>(4,a,b,30)</math>, <math>(4,a,b,50)</math>, <math>(5,a,b,30)</math> and <math>(5,a,b,40)</math> will never be arithmetic, since that would require <math>a</math> and <math>b</math> to be non-integers.
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So, we need to subtract off <math>3</math> progressions from the <math>231</math> we counted, to get our final answer of <math>\boxed{228}</math>.
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~ ihatemath123
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==See Also==
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{{AIME box|year=2022|n=I|num-b=2|num-a=4}}
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{{MAA Notice}}

Revision as of 20:45, 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 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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