Difference between revisions of "2023 AIME II Problems/Problem 15"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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+ | == See also == | ||
+ | {{AIME box|year=2023|num-b=14|after=Last Problem|n=II}} | ||
+ | {{MAA Notice}} |
Revision as of 17:33, 16 February 2023
For each positive integer let
be the least positive integer multiple of
such that
Find the number of positive integers
less than or equal to
that satisfy
Solution
Denote .
Thus, for each
, we need to find smallest positive integer
, such that
Thus, we need to find smallest , such that
Now, we find the smallest , such that
.
We must have
. That is,
.
We find
.
Therefore, for each , we need to find smallest
, such that
We have the following results:
\begin{enumerate}
\item If , then
and
.
\item If
, then
and
.
\item If
, then
and
.
\item If
, then
and
.
\item If
, then
and
.
\item If
, then
and
.
\item If
, then
and
.
\item If
, then
and
.
\item If
, then
and
.
\item If
, then
and
.
\item If
, then
and
.
\end{enumerate}
Therefore, in each cycle, , we have
,
,
,
, such that
. That is,
.
At the boundary of two consecutive cycles,
.
We have .
Therefore, the number of feasible
is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
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.