Difference between revisions of "1980 AHSME Problems/Problem 17"
(Created page with "== Problem == Given that <math>i^2=-1</math>, for how many integers <math>n</math> is <math>(n+i)^4</math> an integer? <math>\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qqu...") |
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== Solution == | == Solution == | ||
− | <math>\ | + | <math>(n+i)^4=n^4+4in^3-6n^2-4in+1</math>, and this has to be an integer, so the sum of the imaginary parts must be <math>0</math>. <cmath>4in^3-4in=0</cmath> <cmath> 4in^3=4in</cmath> <cmath>n^3=n</cmath> |
+ | Since <math>n^3=n</math>, there are <math>\boxed{3}</math> solutions for <math>n</math>, <math>0</math> and <math>\pm1</math>. | ||
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+ | -aopspandy | ||
== See also == | == See also == |
Revision as of 19:09, 18 June 2021
Problem
Given that , for how many integers is an integer?
Solution
, and this has to be an integer, so the sum of the imaginary parts must be . Since , there are solutions for , and .
-aopspandy
See also
1980 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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All AHSME Problems and Solutions |
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