Difference between revisions of "2025 AIME II Problems/Problem 13"

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<cmath>x_{k+1}=\frac{1}{3}\left(x_k+\frac{1}{x_k}-1\right).</cmath><math>x_{2025}</math> can be expressed as <math>\frac{m}{n}</math> for relatively prime positive integers <math>m</math> and <math>n</math>. Find the remainder when <math>m+n</math> is divided by <math>1000</math>.
 
<cmath>x_{k+1}=\frac{1}{3}\left(x_k+\frac{1}{x_k}-1\right).</cmath><math>x_{2025}</math> can be expressed as <math>\frac{m}{n}</math> for relatively prime positive integers <math>m</math> and <math>n</math>. Find the remainder when <math>m+n</math> is divided by <math>1000</math>.
  
== Solution ==
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== Solution 1(Recursion) ==
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Note that <math>x_{k+1} = \frac{1}{3}</math>(<math>\frac{(x_k)^{2} - x_k + 1}{x_k}</math>). An astute reader might recognize the top part as one part of a sum of cubes. I multiplied the entire expression by <math>x_k + 1</math>, moved things around a bit, simplified, and was left with the following generalization:
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<math>x_{k+1} = \frac{(x_k)^{3} + 1}{3x_k(x_k + 1)}</math>. Now, we do the following:
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Set <math>x_k = \frac{m_k}{n_k}</math>. Therefore, <math>x_{k+1} = \frac{m_{k+1}}{n_{k+1}}</math>. We plug these expressions into the <math>x_k</math> and <math>x_{k+1}</math> and simplify to get: <math>\frac{m_{k+1}}{n_{k+1}} = \frac{(m_k)^{3} + (n_k)^{3}}{3(m_k)(n_k)(m_k + n_k)}</math>. Now, as we are looking for the sum of the numerators and denominators of <math>x_2025</math>, 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 <math>m_k + n_k</math> from the numerator and denominator leaves us with <math>\frac{m_{k+1}}{n_{k+1}} = \frac{(m_k)^{2} - (m_k)(n_k) + (n_k)^{2}}{3(m_k)(n_k)}</math>. Now, adding the numerator and denominator as well as keeping the extra factor of <math>3</math>, we get: 3(<math>m_{k+1} + n_{k+1}) = (m_k)^{2} + 2(m_k)(n_k) + (n_k)^{2}</math>. Nicely, we get the recursion that <math>m_{k+1} + n_{k+1} = \frac{(m_k + n_k)^{2}}{3}</math>. Now, by listing out terms using this recursion and doing mod(1000), we get our answer of <math>\boxed{248}</math>.

Revision as of 00:35, 14 February 2025

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 $\boxed{248}$.