Difference between revisions of "2016 AIME I Problems/Problem 7"
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We consider two cases: | We consider two cases: | ||
| − | Case 1: <math>ab \ge -2016</math> | + | Case 1: <math>ab \ge -2016</math> |
| + | |||
| + | In this case, if | ||
<cmath>0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = -\frac{\sqrt{|a+b|}}{ab+100}</cmath> | <cmath>0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = -\frac{\sqrt{|a+b|}}{ab+100}</cmath> | ||
then <math>ab \ne -100</math> and <math>|a + b| = 0 = a + b</math>. Thus <math>ab = -a^2</math> so <math>a^2 < 2016</math>. Thus <math>a = -44,-43, ... , -1, 0, 1, ..., 43, 44</math>, yielding <math>89</math> values. However since <math>ab = -a^2 \ne -100</math>, we have <math>a \ne \pm 10</math>. Thus there are <math>87</math> allowed tuples <math>(a,b)</math> in this case. | then <math>ab \ne -100</math> and <math>|a + b| = 0 = a + b</math>. Thus <math>ab = -a^2</math> so <math>a^2 < 2016</math>. Thus <math>a = -44,-43, ... , -1, 0, 1, ..., 43, 44</math>, yielding <math>89</math> values. However since <math>ab = -a^2 \ne -100</math>, we have <math>a \ne \pm 10</math>. Thus there are <math>87</math> allowed tuples <math>(a,b)</math> in this case. | ||
| − | Case 2: <math>ab \le -2016</math> | + | Case 2: <math>ab \le -2016</math> |
| − | <cmath>0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = \frac{\sqrt{ab+2016} - \sqrt{|a+b|}}{ab+100}</cmath> | + | |
| + | In this case, we want | ||
| + | <cmath>0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = \frac{\sqrt{ab+2016} - \sqrt{|a+b|}i}{ab+100}</cmath> | ||
Squaring, we have the equations <math>ab \ne -100</math> (which always holds in this case) and | Squaring, we have the equations <math>ab \ne -100</math> (which always holds in this case) and | ||
| − | <cmath>ab + 2016 = |a + b|</cmath> | + | <cmath>ab + 2016 = -|a + b|</cmath> |
== See also == | == See also == | ||
{{AIME box|year=2016|n=I|num-b=6|num-a=8}} | {{AIME box|year=2016|n=I|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:12, 4 March 2016
Problem
For integers 
and 
consider the complex number
![\[\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i\]](http://latex.artofproblemsolving.com/9/4/a/94aed5cb77ab692adad418e0f67918ea8fe39dc6.png)
Find the number of ordered pairs of integers 
such that this complex number is a real number.
Solution
We consider two cases:
Case 1: 
In this case, if
![\[0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = -\frac{\sqrt{|a+b|}}{ab+100}\]](http://latex.artofproblemsolving.com/e/4/b/e4bf248d3b281431410bee79a2982a80ace108fc.png)
then 
and 
. Thus 
so 
. Thus 
, yielding 
values. However since 
, we have 
. Thus there are 
allowed tuples in this case.
Case 2: 
In this case, we want
![\[0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = \frac{\sqrt{ab+2016} - \sqrt{|a+b|}i}{ab+100}\]](http://latex.artofproblemsolving.com/7/d/a/7da2f848a2b26d7dc59fe20b3d345736acee963d.png)
Squaring, we have the equations (which always holds in this case) and
![\[ab + 2016 = -|a + b|\]](http://latex.artofproblemsolving.com/5/8/6/586544989f38a62b227f80b4e601c0d9c76a4e96.png)
See also
| 2016 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
