Difference between revisions of "1957 AHSME Problems/Problem 42"
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− | == Problem | + | == Problem == |
If <math>S = i^n + i^{-n}</math>, where <math>i = \sqrt{-1}</math> and <math>n</math> is an integer, then the total number of possible distinct values for <math>S</math> is: | If <math>S = i^n + i^{-n}</math>, where <math>i = \sqrt{-1}</math> and <math>n</math> is an integer, then the total number of possible distinct values for <math>S</math> is: | ||
− | <math>\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ \text{more than 4} </math> | + | <math>\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ \text{more than 4} </math> |
==Solution 1== | ==Solution 1== |
Latest revision as of 09:13, 27 July 2024
Contents
Problem
If , where and is an integer, then the total number of possible distinct values for is:
Solution 1
We first use the fact that . Note that and , so and are periodic with periods at most 4. Therefore, it suffices to check for .
For , we have .
For , we have .
For , we have .
For , we have .
Hence, the answer is .
Solution 2
Notice that the powers of cycle in cycles of 4. So let's see if is periodic.
For : we have .
For : we have .
For : we have .
For : we have .
For : we have again. Well, it can be seen that cycles in periods of 4. Select .
~hastapasta
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 41 |
Followed by Problem 43 | |
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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
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