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 "1971 AHSME Problems/Problem 30"

(created solution page)
 
 
Line 10: Line 10:
  
 
== Solution 1 ==
 
== Solution 1 ==
−
<math>\boxed{\textbf{(D) }\frac{1}{1-x}}</math>.
+
Extend the definition of <math>f_n(x)</math> to <math>n\leq0</math>: Let <math>f_{n-1}(x)</math> be the function such that <math>f_1(f_{n-1}(x))=f_n(x)</math>. From the problem, <math>f_5(x)=f_{35}(x)</math>, so the functions <math>f_1(x),f_2(x),\ldots</math> must repeat in a cycle whose length is a cycle which is a divisor of <math>35-5=30</math>. Thus, if <math>f_a(x)=f_b(x)</math> for integers <math>a</math> and <math>b</math>, we know that <math>a\equiv b</math> [[modulo]] <math>30</math>. Thus, because <math>28\equiv-2\pmod{30}</math>, we know that <math>f_{-2}(x)=f_{28}(x)</math>.
 +
 
 +
It is clear that <math>f_0(x)=x</math>, because <math>f_1(f_0(x))=f_1(x)</math>.
 +
 
 +
Let <math>f_{-1}(x)=y</math>. Then, we know that <math>f_1(y)=f_0(x)=x</math>, so we have the following equation we can solve for <math>y</math>:
 +
\begin{align*}
 +
\frac{2y-1}{y+1} &= x \\
 +
2y-1 &= xy+x \\
 +
y(2-x) &= x+1 \\
 +
y &= \frac{x+1}{2-x}
 +
\end{align*}
 +
 
 +
Let <math>f_{-2}(x)=z</math>. Then, we know that <math>f_1(z)=f_{-1}(x)=\frac{x+1}{2-x}</math>, so we have the following equation we can solve for <math>z</math>:
 +
\begin{align*}
 +
\frac{2z-1}{z+1} &= \frac{x+1}{2-x} \\
 +
4z-2-2xz+x &= xz+x+z+1 \\
 +
3z-3xz &= 3 \\
 +
z(1-x) &= 1 \\
 +
z &= \frac1{1-x}
 +
\end{align*}
 +
 
 +
We derived earlier the fact that <math>z=f_{-2}(x)=f_{28}(x)</math>, so <math>f_{28}(x)=\boxed{\textbf{(D) }\frac{1}{1-x}}</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==
−
<math>\boxed{\textbf{(D) }\frac{1}{1-x}}</math>.
+
Keep solving for the next function in the sequence of <math>f_n(x)</math> while being sure not to make silly algebra mistakes. This process reveals that <math>f_6(x)=x</math>, so <math>f_7(x)=f_1(x)</math>, and the cycle of functions repeats [[modulo]] <math>6</math>. Because <math>28\equiv4\pmod6</math>, we know that <math>f_{28}(x)=f_4(x)=\boxed{\textbf{(D) }\frac{1}{1-x}}</math>, which we calculated on the way to deducing that <math>f_6(x)=x</math>.
  
 
== See Also ==
 
== See Also ==
 
{{AHSME 35p box|year=1971|num-b=29|num-a=31}}
 
{{AHSME 35p box|year=1971|num-b=29|num-a=31}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:53, 8 August 2024

Problem

Given the linear fractional transformation of $x$ into $f_1(x)=\dfrac{2x-1}{x+1}$. Define $f_{n+1}(x)=f_1(f_n(x))$ for $n=1,2,3,\cdots$. Assuming that $f_{35}(x)=f_5(x)$, it follows that $f_{28}(x)$ is equal to

$\textbf{(A) }x\qquad \textbf{(B) }\frac{1}{x}\qquad \textbf{(C) }\frac{x-1}{x}\qquad \textbf{(D) }\frac{1}{1-x}\qquad \textbf{(E) }\text{None of these}$

Solution 1

Extend the definition of $f_n(x)$ to $n\leq0$: Let $f_{n-1}(x)$ be the function such that $f_1(f_{n-1}(x))=f_n(x)$. From the problem, $f_5(x)=f_{35}(x)$, so the functions $f_1(x),f_2(x),\ldots$ must repeat in a cycle whose length is a cycle which is a divisor of $35-5=30$. Thus, if $f_a(x)=f_b(x)$ for integers $a$ and $b$, we know that $a\equiv b$ modulo $30$. Thus, because $28\equiv-2\pmod{30}$, we know that $f_{-2}(x)=f_{28}(x)$.

It is clear that $f_0(x)=x$, because $f_1(f_0(x))=f_1(x)$.

Let $f_{-1}(x)=y$. Then, we know that $f_1(y)=f_0(x)=x$, so we have the following equation we can solve for $y$: \begin{align*} \frac{2y-1}{y+1} &= x \\ 2y-1 &= xy+x \\ y(2-x) &= x+1 \\ y &= \frac{x+1}{2-x} \end{align*}

Let $f_{-2}(x)=z$. Then, we know that $f_1(z)=f_{-1}(x)=\frac{x+1}{2-x}$, so we have the following equation we can solve for $z$: \begin{align*} \frac{2z-1}{z+1} &= \frac{x+1}{2-x} \\ 4z-2-2xz+x &= xz+x+z+1 \\ 3z-3xz &= 3 \\ z(1-x) &= 1 \\ z &= \frac1{1-x} \end{align*}

We derived earlier the fact that $z=f_{-2}(x)=f_{28}(x)$, so $f_{28}(x)=\boxed{\textbf{(D) }\frac{1}{1-x}}$.

Solution 2

Keep solving for the next function in the sequence of $f_n(x)$ while being sure not to make silly algebra mistakes. This process reveals that $f_6(x)=x$, so $f_7(x)=f_1(x)$, and the cycle of functions repeats modulo $6$. Because $28\equiv4\pmod6$, we know that $f_{28}(x)=f_4(x)=\boxed{\textbf{(D) }\frac{1}{1-x}}$, which we calculated on the way to deducing that $f_6(x)=x$.

See Also

1971 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Problem 31
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME 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=1971_AHSME_Problems/Problem_30&oldid=224758"