Difference between revisions of "1980 AHSME Problems/Problem 17"
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== Solution == | == Solution == | ||
<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> | <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> | + | Since <math>n^3=n</math>, there are <math>\boxed{3}</math> solutions for <math>n</math>: <math>0</math> and <math>\pm1</math>. |
-aopspandy | -aopspandy | ||
Revision as of 18:10, 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 
. ![\[4in^3-4in=0\]](http://latex.artofproblemsolving.com/a/8/a/a8af414cb2f71cd27571a9c0205b5820ce123ab4.png)
![\[4in^3=4in\]](http://latex.artofproblemsolving.com/e/2/6/e26aa3fbab18bc8e36430fd9476b5dc37496076e.png)
![\[n^3=n\]](http://latex.artofproblemsolving.com/2/9/b/29b010ba98c3cfac231e971a791e1eaa4c4a0a16.png)
Since 
, there are 
solutions for : and 
.
-aopspandy
See also
| 1980 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
