Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(full solution)
(→‎Solution 2 (Faster))
 
(8 intermediate revisions by 3 users not shown)
Line 19: Line 19:
 
We now need an explicit formula for <math>z_k</math>. We can first experiment with the recursive formula:
 
We now need an explicit formula for <math>z_k</math>. We can first experiment with the recursive formula:
  
−
<math>z_k=\frac{1}{3}z_{k-1}^2</math>
+
<math>z_k=\frac{1}{3}z_{k-1}^2=\frac{1}{3}\left(\frac{1}{3}z_{k-2}^2\right)^2=\frac{1}{3}\left(\frac{1}{3}\left(\frac{1}{3}z_{k-3}^2\right)^2\right)^2</math>
−
 
 
−
<math>=\frac{1}{3}\left(\frac{1}{3}z_{k-2}^2\right)^2</math>
 
−
 
 
−
<math>=\frac{1}{3}\left(\frac{1}{3}\left(\frac{1}{3}z_{k-3}^2\right)^2\right)^2</math>
 
  
 
Notice that the inner <math>\frac{1}{3}</math> is acted upon by two consecutive powers of <math>2</math>. This means that it contributes <math>\left(\left(\frac{1}{3}\right)^2\right)^2=\left(\frac{1}{3}\right)^4</math> to the value of <math>z_k</math>. The next innermost <math>\frac{1}{3}</math> is acted upon by one power of <math>2</math>, so it contributes <math>\left(\frac{1}{3}\right)^2</math> to the value of <math>z_k</math>. Finally, the outermost <math>\frac{1}{3}</math> is acted upon by no powers of <math>2</math>, so it contributes <math>\left(\frac{1}{3}\right)^1</math> to the value of <math>z_k</math>. The overall power of <math>\frac{1}{3}</math> in <math>z_k</math> in terms of <math>z_{k-3}</math> is then <math>4+2+1=2^2+2^1+2^0=2^3-1</math>. Then, the overall power of <math>\frac{1}{3}</math> in <math>z_k</math> in terms of <math>z_{k-a}</math> is <math>2^a-1</math> for positive integers <math>a</math>.
 
Notice that the inner <math>\frac{1}{3}</math> is acted upon by two consecutive powers of <math>2</math>. This means that it contributes <math>\left(\left(\frac{1}{3}\right)^2\right)^2=\left(\frac{1}{3}\right)^4</math> to the value of <math>z_k</math>. The next innermost <math>\frac{1}{3}</math> is acted upon by one power of <math>2</math>, so it contributes <math>\left(\frac{1}{3}\right)^2</math> to the value of <math>z_k</math>. Finally, the outermost <math>\frac{1}{3}</math> is acted upon by no powers of <math>2</math>, so it contributes <math>\left(\frac{1}{3}\right)^1</math> to the value of <math>z_k</math>. The overall power of <math>\frac{1}{3}</math> in <math>z_k</math> in terms of <math>z_{k-3}</math> is then <math>4+2+1=2^2+2^1+2^0=2^3-1</math>. Then, the overall power of <math>\frac{1}{3}</math> in <math>z_k</math> in terms of <math>z_{k-a}</math> is <math>2^a-1</math> for positive integers <math>a</math>.
Line 29: Line 25:
 
We also see that the <math>z_{k-3}</math> term is acted upon by <math>3</math> powers of <math>2</math>, meaning that its power is <math>2\cdot2\cdot2=2^3</math>. We can generalize this, so some <math>z_{k-a}</math> term's power is then <math>2^a</math>.
 
We also see that the <math>z_{k-3}</math> term is acted upon by <math>3</math> powers of <math>2</math>, meaning that its power is <math>2\cdot2\cdot2=2^3</math>. We can generalize this, so some <math>z_{k-a}</math> term's power is then <math>2^a</math>.
  
−
If we combine the above, we obtain the formula <math>z_k=\left(\frac{1}{3}\right)^{2^{a-1}}z_{k-a}^{2^a}</math>. Setting <math>k=a+1</math> results in
+
If we combine the above, we obtain the formula <math>z_k=\left(\frac{1}{3}\right)^{2^a-1}z_{k-a}^{2^a}</math>. Setting <math>k=a+1</math> results in
−
<cmath>z_{a+1}=\left(\frac{1}{3}\right)^{2^{a-1}}z_1^{2^a}=36^{2^a}\cdot\left(\frac{1}{3}\right)^{2^{a-1}}</cmath>
+
<cmath>z_{a+1}=\left(\frac{1}{3}\right)^{2^a-1}z_1^{2^a}=36^{2^a}\cdot\left(\frac{1}{3}\right)^{2^a-1}</cmath>
 
We can simplify this, noting that <math>36^{2^a}=12^{2^a}\cdot3^{2^a}</math>:
 
We can simplify this, noting that <math>36^{2^a}=12^{2^a}\cdot3^{2^a}</math>:
−
<cmath>z_{a+1}=12^{2^a}\cdot3^{2^a}\cdot\left(\frac{1}{3}\right)^{2^{a-1}}=3\cdot12^{2^a}</cmath>
+
<cmath>z_{a+1}=12^{2^a}\cdot3^{2^a}\cdot\left(\frac{1}{3}\right)^{2^a-1}=3\cdot12^{2^a}</cmath>
−
Finally, decrementing <math>a+1</math> to <math>a</math> gives us our recursive equation:
+
Finally, decrementing <math>a+1</math> to <math>a</math> gives us our explicit equation:
 
<cmath>z_a=3\cdot12^{2^{a-1}}</cmath>
 
<cmath>z_a=3\cdot12^{2^{a-1}}</cmath>
  
Line 55: Line 51:
 
== Solution 2 (Faster) ==
 
== Solution 2 (Faster) ==
  
−
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:
+
Note that <math>x_{k+1} = \frac{1}{3}\left( \frac{(x_k)^{2} - x_k + 1}{x_k} \right)</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:
 
<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>.
+
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>.
  
 
~ilikemath247365
 
~ilikemath247365
Line 63: Line 59:
  
 
Note: this works, but why are we able to add the numerator and denominator? What if one of the fractions is not fully simplified?
 
Note: this works, but why are we able to add the numerator and denominator? What if one of the fractions is not fully simplified?
 +
 +
Answer: Because <math>m_k</math> and <math>n_k</math> are coprime, any prime factor <math>p \mid m_k</math> can not be a factor of <math>(n_k)^{2}</math>, any prime factor <math>q \mid n_k</math> can not be a factor of <math>(m_k)^{2}</math>. The only possible common factor of the numerator and denominator could be 3. However, a simple induction argument shows <math>n_k \equiv 2 \mod 3</math> and <math>m_k \equiv 1 \mod 3</math> for <math>k > 1</math>. So this double recursion <math>\frac{m_{k+1}}{n_{k+1}} = \frac{\frac{1}{3}\big((m_k)^{2} - (m_k)(n_k) + (n_k)^{2}\big)}{(m_k)(n_k)}</math> is indeed the most simplified one.
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Rakko12 Rakko12]
  
 
==See also==
 
==See also==

Latest revision as of 22:34, 15 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 (complete)

This problem can be split into three parts, listed below:

Part 1: Analyzing Fractions

Let $x_k=\cfrac{a_k}{b_k}$, where $a_k,b_k$ are relatively prime positive integers. First, we analyze the moduli of the problem. Plugging in for $x_2$ yields $x_2=\frac{157}{275}$. Notice that in both $x_1$ and $x_2$, the numerator is equivalent to $1$ and the denominator is equivalent to $2$ modulus $3$. We see that $x_2=\frac{1}{3}\cdot\frac{(a_1-b_1)^2+a_1b_1}{a_1b_1}$. Specifically, we know that \[(a_1-b_1)^2+a_1b_1\equiv(1-2)^2+1\cdot2\equiv0\hspace{2mm}(\text{mod}\hspace{1mm}3)\] Then this is always divisible by $3$ for all $x_k$ (it can be shown that for all $x_k$, we have $a_k\equiv1\hspace{2mm}(\text{mod}\hspace{1mm}3)$ and $b_k\equiv2\hspace{2mm}(\text{mod}\hspace{1mm}3)$ by using $\text{mod}\hspace{1mm}9$). Thus, $x_2=\frac{\frac{1}{3}((a_1-b_1)^2+a_1b_1)}{a_1b_1}$, and the numerator and denominator of the right-hand side (RHS) correspond to the numerator and denominator of $x_2$ in simplest form. (To further prove that the top and bottom are relatively prime, consider that $a_k$ and $b_k$ are by definition relatively prime, so $(a_k-b_k)^2$ and $a_kb_k$ share no factors.)

Notice that the above do not just apply to $x_1$; we did not use any specific properties of $x_1$. Then we may generalize the above, finding that: \[a_k=\frac{1}{3}((a_{k-1}-b_{k-1})^2+a_{k-1}b_{k-1})\] \[b_k=a_{k-1}b_{k-1}\] Summing the equations yields $a_k+b_k=\frac{1}{3}(a_{k-1}+b_{k-1})^2$ after some manipulation. Let $z_k=a_k+b_k$; then $z_k=\frac{1}{3}z_{k-1}^2$. We are tasked with finding $z_{2025}$.

Part 2: Recursion

We now need an explicit formula for $z_k$. We can first experiment with the recursive formula:

$z_k=\frac{1}{3}z_{k-1}^2=\frac{1}{3}\left(\frac{1}{3}z_{k-2}^2\right)^2=\frac{1}{3}\left(\frac{1}{3}\left(\frac{1}{3}z_{k-3}^2\right)^2\right)^2$

Notice that the inner $\frac{1}{3}$ is acted upon by two consecutive powers of $2$. This means that it contributes $\left(\left(\frac{1}{3}\right)^2\right)^2=\left(\frac{1}{3}\right)^4$ to the value of $z_k$. The next innermost $\frac{1}{3}$ is acted upon by one power of $2$, so it contributes $\left(\frac{1}{3}\right)^2$ to the value of $z_k$. Finally, the outermost $\frac{1}{3}$ is acted upon by no powers of $2$, so it contributes $\left(\frac{1}{3}\right)^1$ to the value of $z_k$. The overall power of $\frac{1}{3}$ in $z_k$ in terms of $z_{k-3}$ is then $4+2+1=2^2+2^1+2^0=2^3-1$. Then, the overall power of $\frac{1}{3}$ in $z_k$ in terms of $z_{k-a}$ is $2^a-1$ for positive integers $a$.

We also see that the $z_{k-3}$ term is acted upon by $3$ powers of $2$, meaning that its power is $2\cdot2\cdot2=2^3$. We can generalize this, so some $z_{k-a}$ term's power is then $2^a$.

If we combine the above, we obtain the formula $z_k=\left(\frac{1}{3}\right)^{2^a-1}z_{k-a}^{2^a}$. Setting $k=a+1$ results in \[z_{a+1}=\left(\frac{1}{3}\right)^{2^a-1}z_1^{2^a}=36^{2^a}\cdot\left(\frac{1}{3}\right)^{2^a-1}\] We can simplify this, noting that $36^{2^a}=12^{2^a}\cdot3^{2^a}$: \[z_{a+1}=12^{2^a}\cdot3^{2^a}\cdot\left(\frac{1}{3}\right)^{2^a-1}=3\cdot12^{2^a}\] Finally, decrementing $a+1$ to $a$ gives us our explicit equation: \[z_a=3\cdot12^{2^{a-1}}\]

Part 3: Mod Bash

Noting that $z_{2025}=3\cdot12^{2^{2024}}$, we are asked to find its value mod $1000$. We can split mod $1000$ into mod $125$ and mod $8$. We know that $z_{2025}\equiv0\hspace{2mm}(\text{mod}\hspace{1mm}8)$, so we must find its value mod $125$.

We find that $\phi(125)=100$, so by Euler's Totient Theorem we know that $12^{100}\equiv1\hspace{2mm}(\text{mod}\hspace{1mm}125)$. Then, since the power of $12$ is $2^{2024}$, we must find this over modulus $100$.

Again, we split mod $100$ into mod $4$ and mod $25$. We know that $2^{2024}\equiv0\hspace{2mm}(\text{mod}\hspace{1mm}4)$. Since $\phi(25)=20$, we can apply Euler again, finding that $2^{20}\equiv1\hspace{2mm}(\text{mod}\hspace{1mm}25)$. Then \[2^{2024}\equiv(2^{20})^{101}\cdot2^4\equiv2^4\equiv16\hspace{2mm}(\text{mod}\hspace{1mm}25)\] Combining this with the mod $4$ result yields $2^{2024}\equiv16\hspace{2mm}(\text{mod}\hspace{1mm}100)$.

Going back, $12^{2^{2024}}\equiv12^{16}\hspace{2mm}(\text{mod}\hspace{1mm}125)$. We can then decrement this using a series of simplifications: \[12^{16}\equiv144^8\equiv19^8\equiv361^4\equiv(-14)^4\equiv196^2\equiv(-54)^2\equiv2916\equiv41\hspace{2mm}(\text{mod}\hspace{1mm}125)\] Remember that the original value of $z_{2025}$ included a multiplication of $3$; thus, \[z_{2025}\equiv41\cdot3\equiv123\hspace{2mm}(\text{mod}\hspace{1mm}125)\] Finally, combining this with the fact that $z_{2025}\equiv0\hspace{2mm}(\text{mod}\hspace{1mm}8)$, we find that the solution to the system of moduli is $z_{2025}\equiv\boxed{248}\hspace{2mm}(\text{mod}\hspace{1mm}1000)$.

~eevee9406

Solution 2 (Faster)

Note that $x_{k+1} = \frac{1}{3}\left( \frac{(x_k)^{2} - x_k + 1}{x_k} \right)$. 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


Note: this works, but why are we able to add the numerator and denominator? What if one of the fractions is not fully simplified?

Answer: Because $m_k$ and $n_k$ are coprime, any prime factor $p \mid m_k$ can not be a factor of $(n_k)^{2}$, any prime factor $q \mid n_k$ can not be a factor of $(m_k)^{2}$. The only possible common factor of the numerator and denominator could be 3. However, a simple induction argument shows $n_k \equiv 2 \mod 3$ and $m_k \equiv 1 \mod 3$ for $k > 1$. So this double recursion $\frac{m_{k+1}}{n_{k+1}} = \frac{\frac{1}{3}\big((m_k)^{2} - (m_k)(n_k) + (n_k)^{2}\big)}{(m_k)(n_k)}$ is indeed the most simplified one.

~Rakko12

See also

2025 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2025_AIME_II_Problems/Problem_13&oldid=242859"