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 == | + | == Solution 1(Recursion) == |
| + | |||
| + | 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: | ||
| + | <math>x_{k+1} = \frac{(x_k)^{3} + 1}{3x_k(x_k + 1)}</math>. Now, we do the following: | ||
| + | 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 
be defined such that 
and
![\[x_{k+1}=\frac{1}{3}\left(x_k+\frac{1}{x_k}-1\right).\]](http://latex.artofproblemsolving.com/4/d/6/4d6ef61e00711b8fadd71ef58b4b6ec2b6fa1c6b.png)
can be expressed as 
for relatively prime positive integers 
and 
. Find the remainder when 
is divided by 
.
Solution 1(Recursion)
Note that 
(
). An astute reader might recognize the top part as one part of a sum of cubes. I multiplied the entire expression by 
, moved things around a bit, simplified, and was left with the following generalization:
. Now, we do the following:
Set 
. Therefore, 
. We plug these expressions into the 
and 
and simplify to get: 
. Now, as we are looking for the sum of the numerators and denominators of 
, 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 
from the numerator and denominator leaves us with 
. Now, adding the numerator and denominator as well as keeping the extra factor of 
, we get: 3(
. Nicely, we get the recursion that 
. Now, by listing out terms using this recursion and doing mod(1000), we get our answer of 
.
