Difference between revisions of "1957 AHSME Problems/Problem 42"
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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> | + | 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>. |
Latest revision as of 14:47, 10 June 2024
Problem 42
If , where
and
is an integer, then the total number of possible distinct values for
is:
Solution
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