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Difference between revisions of "2016 AIME I Problems/Problem 7"

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<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>
 
<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>
 
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  
−
<math></math>ab + 2016 = |a + b|
+
<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 18:09, 4 March 2016

Problem

For integers $a$ and $b$ consider the complex number \[\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i\]

Find the number of ordered pairs of integers $(a,b)$ such that this complex number is a real number.

Solution

We consider two cases:

Case 1: $ab \ge -2016$ In this case, if \[0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = -\frac{\sqrt{|a+b|}}{ab+100}\] then $ab \ne -100$ and $|a + b| = 0 = a + b$. Thus $ab = -a^2$ so $a^2 < 2016$. Thus $a = -44,-43, ... , -1, 0, 1, ..., 43, 44$, yielding $89$ values. However since $ab = -a^2 \ne -100$, we have $a \ne \pm 10$. Thus there are $87$ allowed tuples $(a,b)$ in this case.

Case 2: $ab \le -2016$. In this case, we want \[0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = \frac{\sqrt{ab+2016} - \sqrt{|a+b|}}{ab+100}\] Squaring, we have the equations $ab \ne -100$ (which always holds in this case) and \[ab + 2016 = |a + b|\]

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

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