Difference between revisions of "2022 AIME I Problems/Problem 6"
(→Solution 2) |
(→Solution 2) |
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Line 33: | Line 33: | ||
Hence, this problem asks us to compute | Hence, this problem asks us to compute | ||
− | < | + | <cmath> |
\[ | \[ | ||
| S | - \left( | A | + | B | + | C | \right) . | | S | - \left( | A | + | B | + | C | \right) . | ||
\] | \] | ||
− | </ | + | </cmath> |
First, we compute <math>| S |</math>. | First, we compute <math>| S |</math>. | ||
Line 45: | Line 45: | ||
Second, we compute <math>| A |</math>. | Second, we compute <math>| A |</math>. | ||
− | \textbf{Case 1}: <math>a = 6</math>. | + | <math>\textbf{Case 1}</math>: <math>a = 6</math>. |
We have <math>b = 8 , \cdots , 19, 21, 22, \cdots, 29</math>. | We have <math>b = 8 , \cdots , 19, 21, 22, \cdots, 29</math>. | ||
Thus, the number of solutions is 21. | Thus, the number of solutions is 21. | ||
− | \textbf{Case 2}: <math>a = 20</math>. | + | <math>\textbf{Case 2}</math>: <math>a = 20</math>. |
We have <math>b = 21, 22, \cdots , 29</math>. | We have <math>b = 21, 22, \cdots , 29</math>. | ||
Line 67: | Line 67: | ||
Fourth, we compute <math>| C |</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>. | + | <math>\textbf{Case 1}</math>: 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>. | Hence, <math>(a, b) = (6 , 7), (7, 9) , (10, 20)</math>. | ||
Therefore, the number solutions in this case is 3. | 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>. | + | <math>\textbf{Case 2}</math>: 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>. | + | <math>\textbf{Case 2.1}</math>: The arithmetic sequence is <math>3, a, b, 30</math>. |
Hence, <math>(a, b) = (12, 21)</math>. | Hence, <math>(a, b) = (12, 21)</math>. | ||
− | \textbf{Case 2.2}: The arithmetic sequence is <math>4, a, b, 40</math>. | + | <math>\textbf{Case 2.2}</math>: The arithmetic sequence is <math>4, a, b, 40</math>. |
Hence, <math>(a, b) = (16, 28)</math>. | Hence, <math>(a, b) = (16, 28)</math>. | ||
− | \textbf{Case 2.3}: The arithmetic sequence is <math>5, a, b, 50</math>. | + | <math>\textbf{Case 2.3}</math>: The arithmetic sequence is <math>5, a, b, 50</math>. |
Hence, <math>(a, b) = (20, 35)</math>. | Hence, <math>(a, b) = (20, 35)</math>. | ||
Line 89: | Line 89: | ||
Therefore, | Therefore, | ||
+ | <cmath> | ||
\begin{align*} | \begin{align*} | ||
| S | - \left( | A | + | B | + | C | \right) | | S | - \left( | A | + | B | + | C | \right) | ||
Line 94: | Line 95: | ||
& = \boxed{\textbf{(228) }} . | & = \boxed{\textbf{(228) }} . | ||
\end{align*} | \end{align*} | ||
+ | </cmath> | ||
~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) |
Revision as of 22:36, 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
First, we compute .
We have .
Second, we compute .
: .
We have . Thus, the number of solutions is 21.
: .
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 .
: 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.
: In the arithmetic sequence, the two numbers beyond and are on the opposite sides of and .
: The arithmetic sequence is .
Hence, .
: The arithmetic sequence is .
Hence, .
: 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.