Difference between revisions of "1957 AHSME Problems/Problem 42"

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== Problem 42==
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== 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>  
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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>
  
 
==Solution 1==
 
==Solution 1==

Latest revision as of 09:13, 27 July 2024

Problem

If $S = i^n + i^{-n}$, where $i = \sqrt{-1}$ and $n$ is an integer, then the total number of possible distinct values for $S$ is:

$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ \text{more than 4}$

Solution 1

We first use the fact that $i^{-n}=\frac1{i^n}=\left(\frac1i\right)^n=(-i)^n$. Note that $i^4=1$ and $(-i)^4=1$, so $i^n$ and $(-i)^n$ are periodic with periods at most 4. Therefore, it suffices to check for $n=0,1,2,3$.


For $n=0$, we have $i^0+(-i)^0=1+1=2$.

For $n=1$, we have $i^1+(-i)^1=i-i=0$.

For $n=2$, we have $i^2+(-i)^2=-1-1=-2$.

For $n=3$, we have $i^3+(-i)^3=-i+i=0$.

Hence, the answer is $\boxed{\textbf{(C)}\ 3}$.

Solution 2

Notice that the powers of $i$ cycle in cycles of 4. So let's see if $S$ is periodic.

For $n=0$: we have $2$.

For $n=1$: we have $0$.

For $n=2$: we have $-2$.

For $n=3$: we have $0$.

For $n=4$: we have $2$ again. Well, it can be seen that $S$ cycles in periods of 4. Select $\fbox{\textbf{(C)}}$.

~hastapasta

See Also

1957 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 41
Followed by
Problem 43
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All AHSME Problems and Solutions

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