1971 AHSME Problems/Problem 30
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 | |
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