2023 AIME II Problems/Problem 15

For each positive integer $n$ let $a_n$ be the least positive integer multiple of $23$ such that $a_n \equiv 1 \pmod{2^n}.$ Find the number of positive integers $n$ less than or equal to $1000$ that satisfy $a_n = a_{n+1}.$

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)

Solution 2

Observe that if $a_{n-1}$ is divisible by $2^n$ , $a_n = a_{n-1}$ . If not, $a_n = a_{n-1} + 23 \cdot 2^{n-1}$ .

This encourages us to let $b_n = (a_n - 1)/2^n$ . Rewriting the above equations, we have $b_n = \begin{cases} b_{n-1}/2 & \text{if } 2 | (b_{n-1}+23)/2 \\ 1 &\text{if } 2\not\vert b_{n-1} \end{cases}$ Now, we start listing. We know that $b_1 = 11$ , so $b_2 = 17$ , $b_3 = 20$ , $b_4 = 10$ , $b_5 = 5$ , $b_6 = 14$ , $b_7 = 7$ , $b_8 = 15$ , $b_9 = 19$ , $b_{10} = 21$ , $b_{11} = 22$ , $b_{12} = 1$ . Hence the sequence is periodic with period 11. Note that $a_n = a_{n+1}$ if and only if $b_n = b_{n+1}/2$ , i.e. $b_n$ is even. This occurrs when $n$ is congruent to 0, 3, 4 or 6 mod 11.

From 1 to $1001 = 91 \times 11$ , there are $91 \times 4 = 364$ values of $n$ . Since $1001$ satisfies the criteria, we subtract 1 to get $\fbox{363}$ , and we're done!

~Bxiao31415

2023 AIME II (Problems • Answer Key • Resources)
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2023 AIME II Problems/Problem 15

Solution

Solution 2

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