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

(Created page with "== Problem 42== 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 <m...")
 
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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:  
  
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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>  
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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>
  
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==Solution==
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==Solution 1==
  
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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>.
  
  
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Hence, the answer is <math>\boxed{\textbf{(C)}\ 3}</math>.
 
Hence, the answer is <math>\boxed{\textbf{(C)}\ 3}</math>.
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== Solution 2 ==
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Notice that the powers of <math>i</math> cycle in cycles of 4. So let's see if <math>S</math> is periodic.
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For <math>n=0</math>: we have <math>2</math>.
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For <math>n=1</math>: we have <math>0</math>.
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For <math>n=2</math>: we have <math>-2</math>.
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For <math>n=3</math>: we have <math>0</math>.
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For <math>n=4</math>: we have <math>2</math> again. Well, it can be seen that <math>S</math> cycles in periods of 4. Select <math>\fbox{\textbf{(C)}}</math>.
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~hastapasta
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== See Also ==
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{{AHSME 50p box|year=1957|num-b=41|num-a=43}}
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{{MAA Notice}}
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[[Category:AHSME]][[Category:AHSME Problems]]

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 (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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