1971 AHSME Problems/Problem 30
Contents
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 
:
\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 
. Then, we know that 
, so we have the following equation we can solve for 
:
\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 
, 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 | |
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