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 23:36, 17 February 2022
Contents
Problem
Find the number of ordered pairs of integers such that the sequence
is 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.