2023 AIME II Problems/Problem 15
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)