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

(Problem)
(Problem)
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Find the number of ordered pairs of integers <math>(a,b)</math> such that this complex number is a real number.
 
Find the number of ordered pairs of integers <math>(a,b)</math> such that this complex number is a real number.
  
== Solution == We consider two cases:
+
==Solution== We consider two cases:
  
Case 1:  <math>ab > -2016</math>  In this case, if
+
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>,
+
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>.  In this case, we want
 +
<cmath>0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = \frac{\sqrt{|a+b|}}{ab+100}}</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 18:46, 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

\[0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = \frac{\sqrt{|a+b|}}{ab+100}}\] (Error compiling LaTeX. Unknown error_msg)

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

\[0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = \frac{\sqrt{|a+b|}}{ab+100}}\] (Error compiling LaTeX. Unknown error_msg)

See also

2016 AIME I (ProblemsAnswer KeyResources)
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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