Difference between revisions of "2022 AIME I Problems/Problem 6"
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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. | 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. | ||
− | == Solution == | + | == Solution 1== |
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. | 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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~ ihatemath123 | ~ ihatemath123 | ||
+ | |||
+ | ==Solution 2== | ||
+ | Denote <math>S = \left\{ (a, b) : 6 \leq a < b \leq 29 \right\}</math>. | ||
+ | |||
+ | Denote by <math>A</math> a subset of <math>S</math>, such that there exists an arithmetic sequence that has 4 terms and includes <math>a</math> but not <math>b</math>. | ||
+ | |||
+ | Denote by <math>B</math> a subset of <math>S</math>, such that there exists an arithmetic sequence that has 4 terms and includes <math>b</math> but not <math>a</math>. | ||
+ | |||
+ | Hence, <math>C</math> is a subset of <math>S</math>, such that there exists an arithmetic sequence that has 4 terms and includes both <math>a</math> and <math>b</math>. | ||
+ | |||
+ | Hence, this problem asks us to compute | ||
+ | \[ | ||
+ | | S | - \left( | A | + | B | + | C | \right) . | ||
+ | \] | ||
+ | |||
+ | First, we compute <math>| S |</math>. | ||
+ | |||
+ | We have <math>| S | = \binom{29 - 6 + 1}{2} = \binom{24}{2} = 276</math>. | ||
+ | |||
+ | Second, we compute <math>| A |</math>. | ||
+ | |||
+ | \textbf{Case 1}: <math>a = 6</math>. | ||
+ | |||
+ | We have <math>b = 8 , \cdots , 19, 21, 22, \cdots, 29</math>. | ||
+ | Thus, the number of solutions is 21. | ||
+ | |||
+ | \textbf{Case 2}: <math>a = 20</math>. | ||
+ | |||
+ | We have <math>b = 21, 22, \cdots , 29</math>. | ||
+ | Thus, the number of solutions is 9. | ||
+ | |||
+ | Thus, <math>| A | = 21 + 9 = 30</math>. | ||
+ | |||
+ | Third, we compute <math>| B |</math>. | ||
+ | |||
+ | In <math>B</math>, we have <math>b = 6, 20</math>. However, because <math>6 \leq a < b</math>, we have <math>b \geq 7</math>. | ||
+ | Thus, <math>b = 20</math>. | ||
+ | |||
+ | This implies <math>a = 7, 8, 9, 11, 12, \cdots , 19</math>. | ||
+ | Thus, <math>| B | = 12</math>. | ||
+ | |||
+ | Fourth, we compute <math>| C |</math>. | ||
+ | |||
+ | \textbf{Case 1}: In the arithmetic sequence, the two numbers beyond <math>a</math> and <math>b</math> are on the same side of <math>a</math> and <math>b</math>. | ||
+ | |||
+ | Hence, <math>(a, b) = (6 , 7), (7, 9) , (10, 20)</math>. | ||
+ | Therefore, the number solutions in this case is 3. | ||
+ | |||
+ | \textbf{Case 2}: In the arithmetic sequence, the two numbers beyond <math>a</math> and <math>b</math> are on the opposite sides of <math>a</math> and <math>b</math>. | ||
+ | |||
+ | \textbf{Case 2.1}: The arithmetic sequence is <math>3, a, b, 30</math>. | ||
+ | |||
+ | Hence, <math>(a, b) = (12, 21)</math>. | ||
+ | |||
+ | \textbf{Case 2.2}: The arithmetic sequence is <math>4, a, b, 40</math>. | ||
+ | |||
+ | Hence, <math>(a, b) = (16, 28)</math>. | ||
+ | |||
+ | \textbf{Case 2.3}: The arithmetic sequence is <math>5, a, b, 50</math>. | ||
+ | |||
+ | Hence, <math>(a, b) = (20, 35)</math>. | ||
+ | |||
+ | Putting two cases together, <math>| C | = 6</math>. | ||
+ | |||
+ | Therefore, | ||
+ | \begin{align*} | ||
+ | | S | - \left( | A | + | B | + | C | \right) | ||
+ | & = 276 - \left( 30 + 12 + 6 \right) \ | ||
+ | & = \boxed{\textbf{(228) }} . | ||
+ | \end{align*} | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=I|num-b=5|num-a=7}} | {{AIME box|year=2022|n=I|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:34, 17 February 2022
Contents
[hide]Problem
Find the number of ordered pairs of integers such that the sequenceis strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.
Solution 1
Since and cannot be an arithmetic progression, or can never be . Since and cannot be an arithmetic progression, can never be . Since , there are ways to choose and with these two restrictions in mind.
However, there are still specific invalid cases counted in these pairs . Since cannot form an arithmetic progression, . cannot be an arithmetic progression, so ; however, since this pair was not counted in our , we do not need to subtract it off. cannot form an arithmetic progression, so . cannot form an arithmetic progression, so . cannot form an arithmetic progression, ; however, since this pair was not counted in our (since we disallowed or to be ), we do not to subtract it off.
Also, the sequences , , , , and will never be arithmetic, since that would require and to be non-integers.
So, we need to subtract off progressions from the we counted, to get our final answer of .
~ ihatemath123
Solution 2
Denote .
Denote by a subset of , such that there exists an arithmetic sequence that has 4 terms and includes but not .
Denote by a subset of , such that there exists an arithmetic sequence that has 4 terms and includes but not .
Hence, is a subset of , such that there exists an arithmetic sequence that has 4 terms and includes both and .
Hence, this problem asks us to compute \[ | S | - \left( | A | + | B | + | C | \right) . \]
First, we compute .
We have .
Second, we compute .
\textbf{Case 1}: .
We have . Thus, the number of solutions is 21.
\textbf{Case 2}: .
We have . Thus, the number of solutions is 9.
Thus, .
Third, we compute .
In , we have . However, because , we have . Thus, .
This implies . Thus, .
Fourth, we compute .
\textbf{Case 1}: In the arithmetic sequence, the two numbers beyond and are on the same side of and .
Hence, . Therefore, the number solutions in this case is 3.
\textbf{Case 2}: In the arithmetic sequence, the two numbers beyond and are on the opposite sides of and .
\textbf{Case 2.1}: The arithmetic sequence is .
Hence, .
\textbf{Case 2.2}: The arithmetic sequence is .
Hence, .
\textbf{Case 2.3}: The arithmetic sequence is .
Hence, .
Putting two cases together, .
Therefore,
~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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