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Difference between revisions of "1957 AHSME Problems/Problem 42"
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==Solution==
 
==Solution==
  
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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> have are periodic with periods at most 4. Therefore, it suffices to check for <math>n=0,1,2,3</math>.
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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 13:47, 10 June 2024

Problem 42

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

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 $\boxed{C}$.

~hastapasta

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