2023 AIME II Problems/Problem 15
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)