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

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==Problem==
 
==Problem==
 
For integers <math>a</math> and <math>b</math> consider the complex number
 
For integers <math>a</math> and <math>b</math> consider the complex number
−
<cmath>\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i</cmath>
+
<cmath>\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i</cmath>
  
 
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:
 +
 +
'''Case 1:'''  <math>ab \ge -2016</math>. 
 +
 +
In this case, if
 +
<cmath>0 = \text{Im}\left({\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i}\right) = -\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.
 +
 +
'''Case 2:'''  <math>ab < -2016</math>.
 +
 +
In this case, we want
 +
<cmath>0 = \text{Im}\left({\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i}\right) = \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
 +
<cmath>-(ab + 2016)= |a + b|.</cmath>
 +
Then if <math>a > 0</math> and <math>b < 0</math>, let <math>c = -b</math>.  If <math>c > a</math>,
 +
<cmath>ac - 2016 = c - a \Rightarrow (a - 1)(c + 1) = 2015 \Rightarrow (a,b) = (2, -2014), (6, -402), (14, -154), (32, -64). </cmath>
 +
Note that <math>ab < -2016</math> for every one of these solutions.  If <math>c < a</math>, then
 +
<cmath>ac - 2016 = a - c \Rightarrow (a + 1)(c - 1) = 2015 \Rightarrow (a,b) = (2014, -2), (402, -6), (154, -14), (64, -32). </cmath>
 +
Again, <math>ab < -2016</math> for every one of the above solutions. This yields <math>8</math> solutions.  Similarly, if <math>a < 0</math> and <math>b > 0</math>, there are <math>8</math> solutions.  Thus, there are a total of <math>16</math> solutions in this case.
 +
 +
Thus, the answer is <math>87 + 16 = \boxed{103}</math>.
  
 
== 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}}

Latest revision as of 17:00, 2 January 2024

Problem

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

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 \[0 = \text{Im}\left({\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i}\right) = -\frac{\sqrt{|a+b|}}{ab+100}\] 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: $ab < -2016$.

In this case, we want \[0 = \text{Im}\left({\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i}\right) = \frac{\sqrt{-ab-2016} - \sqrt{|a+b|}}{ab+100}\] Squaring, we have the equations $ab \ne -100$ (which always holds in this case) and \[-(ab + 2016)= |a + b|.\] Then if $a > 0$ and $b < 0$, let $c = -b$. If $c > a$, \[ac - 2016 = c - a \Rightarrow (a - 1)(c + 1) = 2015 \Rightarrow (a,b) = (2, -2014), (6, -402), (14, -154), (32, -64).\] Note that $ab < -2016$ for every one of these solutions. If $c < a$, then \[ac - 2016 = a - c \Rightarrow (a + 1)(c - 1) = 2015 \Rightarrow (a,b) = (2014, -2), (402, -6), (154, -14), (64, -32).\] Again, $ab < -2016$ for every one of the above solutions. This yields $8$ solutions. Similarly, if $a < 0$ and $b > 0$, there are $8$ solutions. Thus, there are a total of $16$ solutions in this case.

Thus, the answer is $87 + 16 = \boxed{103}$.

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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