Difference between revisions of "1971 AHSME Problems/Problem 30"
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== 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 15:53, 8 August 2024
Contents
[hide]Problem
Given the linear fractional transformation of into . Define for . Assuming that , it follows that is equal to
Solution 1
Extend the definition of to : Let be the function such that . From the problem, , so the functions must repeat in a cycle whose length is a cycle which is a divisor of . Thus, if for integers and , we know that modulo . Thus, because , we know that .
It is clear that , because .
Let . Then, we know that , so we have the following equation we can solve for :
Let . Then, we know that , so we have the following equation we can solve for :
We derived earlier the fact that , so .
Solution 2
Keep solving for the next function in the sequence of while being sure not to make silly algebra mistakes. This process reveals that , so , and the cycle of functions repeats modulo . Because , we know that , which we calculated on the way to deducing that .
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.