2022 AIME I Problems/Problem 6
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
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
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.