2025 AIME II Problems/Problem 13

Problem

Let the sequence of rationals $x_1,x_2,\dots$ be defined such that $x_1=\frac{25}{11}$ and $x_{k+1}=\frac{1}{3}\left(x_k+\frac{1}{x_k}-1\right).$ $x_{2025}$ can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$ . Find the remainder when $m+n$ is divided by $1000$ .

Solution 1(Recursion)

Note that $x_{k+1} = \frac{1}{3}$ ( $\frac{(x_k)^{2} - x_k + 1}{x_k}$ ). An astute reader might recognize the top part as one part of a sum of cubes. I multiplied the entire expression by $x_k + 1$ , moved things around a bit, simplified, and was left with the following generalization: $x_{k+1} = \frac{(x_k)^{3} + 1}{3x_k(x_k + 1)}$ . Now, we do the following: Set $x_k = \frac{m_k}{n_k}$ . Therefore, $x_{k+1} = \frac{m_{k+1}}{n_{k+1}}$ . We plug these expressions into the $x_k$ and $x_{k+1}$ and simplify to get: $\frac{m_{k+1}}{n_{k+1}} = \frac{(m_k)^{3} + (n_k)^{3}}{3(m_k)(n_k)(m_k + n_k)}$ . Now, as we are looking for the sum of the numerators and denominators of $x_2025$ , this is great! Now, recall that we want the fraction to be simplest. So we have to cancel out anything we can. Canceling out the factor of $m_k + n_k$ from the numerator and denominator leaves us with $\frac{m_{k+1}}{n_{k+1}} = \frac{(m_k)^{2} - (m_k)(n_k) + (n_k)^{2}}{3(m_k)(n_k)}$ . Now, adding the numerator and denominator as well as keeping the extra factor of $3$ , we get: 3( $m_{k+1} + n_{k+1}) = (m_k)^{2} + 2(m_k)(n_k) + (n_k)^{2}$ . Nicely, we get the recursion that $m_{k+1} + n_{k+1} = \frac{(m_k + n_k)^{2}}{3}$ . Now, by listing out terms using this recursion and doing mod(1000), we get our answer of $\boxed{248}$ .

~ilikemath247365

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2025 AIME II Problems/Problem 13

Problem

Solution 1(Recursion)