2023 AIME II Problems/Problem 15

Solution

Denote $a_n = 23 b_n$ . Thus, for each $n$ , we need to find smallest positive integer $k_n$ , such that $23 b_n = 2^n k_n + 1 .$

Thus, we need to find smallest $k_n$ , such that $2^n k_n \equiv - 1 \pmod{23} .$

Now, we find the smallest $m$ , such that $2^m \equiv 1 \pmod{23}$ . We must have $m | \phi \left( 23 \right)$ . That is, $m | 22$ . We find $m = 11$ .

Therefore, for each $n$ , we need to find smallest $k_n$ , such that $2^{{\rm Rem} \left( n , 11 \right)} k_n \equiv - 1 \pmod{23} .$

We have the following results: \begin{enumerate} \item If ${\rm Rem} \left( n , 11 \right) = 0$ , then $k_n = 22$ and $b_n = 1$ . \item If ${\rm Rem} \left( n , 11 \right) = 1$ , then $k_n = 11$ and $b_n = 1$ . \item If ${\rm Rem} \left( n , 11 \right) = 2$ , then $k_n = 17$ and $b_n = 3$ . \item If ${\rm Rem} \left( n , 11 \right) = 3$ , then $k_n = 20$ and $b_n = 7$ . \item If ${\rm Rem} \left( n , 11 \right) = 4$ , then $k_n = 10$ and $b_n = 7$ . \item If ${\rm Rem} \left( n , 11 \right) = 5$ , then $k_n = 5$ and $b_n = 7$ . \item If ${\rm Rem} \left( n , 11 \right) = 6$ , then $k_n = 14$ and $b_n = 39$ . \item If ${\rm Rem} \left( n , 11 \right) = 7$ , then $k_n = 7$ and $b_n = 39$ . \item If ${\rm Rem} \left( n , 11 \right) = 8$ , then $k_n = 15$ and $b_n = 167$ . \item If ${\rm Rem} \left( n , 11 \right) = 9$ , then $k_n = 19$ and $b_n = 423$ . \item If ${\rm Rem} \left( n , 11 \right) = 10$ , then $k_n = 21$ and $b_n = 935$ . \end{enumerate}

Therefore, in each cycle, $n \in \left\{ 11 l , 11l + 1 , \cdots , 11l + 10 \right\}$ , we have $n = 11l$ , $11l + 3$ , $11l + 4$ , $11l + 6$ , such that $b_n = b_{n+1}$ . That is, $a_n = a_{n+1}$ . At the boundary of two consecutive cycles, $b_{11L + 10} \neq b_{11 \left(l + 1 \right)}$ .

We have $1000 = 90 \cdot 11 + 10$ . Therefore, the number of feasible $n$ is $91 \cdot 4 - 1 = \boxed{\textbf{(363) }}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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2023 AIME II Problems/Problem 15

Solution