Difference between revisions of "1957 AHSME Problems/Problem 42"
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| − | == Problem | + | == 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> | + | <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>. | ||
| Line 32: | Line 32: | ||
For <math>n=3</math>: we have <math>0</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>\ | + | 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>. |
~hastapasta | ~hastapasta | ||
| + | |||
| + | == See Also == | ||
| + | {{AHSME 50p box|year=1957|num-b=41|num-a=43}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:AHSME]][[Category:AHSME Problems]] | ||
Latest revision as of 09:13, 27 July 2024
Contents
Problem
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 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
