Difference between revisions of "1957 AHSME Problems/Problem 42"
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<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== | + | ==Solution 1== |
We first use the fact that <math>i^{-n}=\frac1{i^n}=\left(\frac1i\right)^n=(-i)^n</math>. Note that <math>i^4=1</math> and <math>(-i)^4=1</math>, so <math>i^n</math> and <math>(-i)^n</math> are periodic with periods at most 4. Therefore, it suffices to check for <math>n=0,1,2,3</math>. | We first use the fact that <math>i^{-n}=\frac1{i^n}=\left(\frac1i\right)^n=(-i)^n</math>. Note that <math>i^4=1</math> and <math>(-i)^4=1</math>, so <math>i^n</math> and <math>(-i)^n</math> are periodic with periods at most 4. Therefore, it suffices to check for <math>n=0,1,2,3</math>. |
Revision as of 09:12, 27 July 2024
Contents
Problem 42
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 | |
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All AHSME Problems and Solutions |
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