2023 AIME II Problem 15

$a_n = 23 b_n$ where $b_n$ is the inverse of $23$ mod $2^n;$ then $a_n = a_{n+1}$ exactly when the binary digit of weight $2^n$ in $b_{n+1}$ is $0$

Since $23 \cdot 89 = 2^{11} - 1,$ those binary digits repeat with period $11;$ find which four residues of $n$ mod $11$ give a zero digit

Write $a_n = 23 b_n,$ where $b_n$ is the unique integer in $[1, 2^n]$ with $23 b_n \equiv 1 \pmod{2^n},$ i.e. the inverse of $23$ mod $2^n.$ Reducing mod $2^n$ shows $b_{n+1} \in \{b_n,\ b_n + 2^n\},$ so $a_n = a_{n+1}$ exactly when $b_{n+1} = b_n,$ which happens exactly when the binary digit of weight $2^n$ in $b_{n+1}$ is $0.$

Since $23 \cdot 89 = 2047 = 2^{11} - 1,$ setting $T_k = 1 + 2^{11} + 2^{22}$ ${}+ \cdots + 2^{11(k-1)}$ gives $23 \cdot 89\,T_k = 2^{11k} - 1,$ so $$\begin{gathered} 23\left(2^{11k} - 89\,T_k\right) \\ \equiv 1 \pmod{2^{11k}}, \end{gathered}$$ and every $b_n$ with $n \le 11k$ is the reduction of $2^{11k} - 89\,T_k$ mod $2^n.$ In binary, $89 = 1011001_2$ occupies positions $0, 3, 4, 6$ of each $11$-bit block of $89\,T_k.$ Since $89\,T_k$ is odd, $2^{11k} - 89\,T_k$ $= \bigl(2^{11k} - 1 - 89\,T_k\bigr) + 1$ keeps digit $0$ equal to $1$ and complements every digit in positions $1$ through $11k - 1.$

So for $n \ge 1,$ the digit of weight $2^n$ is $0$ exactly when $n \equiv 0, 3, 4, 6 \pmod{11}.$ Among $1 \le n \le 1000 = 90 \cdot 11 + 10,$ the residue $0$ occurs $90$ times and the residues $3,$ $4,$ $6$ occur $91$ times each, for a total of $90 + 3 \cdot 91 = 363.$