Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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...")
 
Line 19: Line 19:
  
 
Hence, the answer is <math>\boxed{\textbf{(C)}\ 3}</math>.
 
Hence, the answer is <math>\boxed{\textbf{(C)}\ 3}</math>.
 +
 +
== Solution 2 ==
 +
 +
Notice that the powers of <math>i</math> cycle in cycles of 4. So let's see if <math>S</math> is periodic.
 +
 +
For <math>n=0</math>: we have <math>2</math>.
 +
 +
For <math>n=1</math>: we have <math>0</math>.
 +
 +
For <math>n=2</math>: we have <math>-2</math>.
 +
 +
For <math>n=3</math>: we have <math>0</math>.
 +
 +
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>\boxed{C}</math>.
 +
 +
~hastapasta

Revision as of 10:35, 2 May 2022

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 $(-i)^n$ have 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1957_AHSME_Problems/Problem_42&oldid=173732"