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

m (Solution)
m (Solution)
Line 9: Line 9:
 
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
 
In this case, if
Line 15: Line 15:
 
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 < -2016</math>.  
+
'''Case 2:''' <math>ab < -2016</math>.  
  
 
In this case, we want
 
In this case, we want
Line 28: Line 28:
  
 
Thus, the answer is <math>87 + 16 = \boxed{103}</math>.
 
Thus, the answer is <math>87 + 16 = \boxed{103}</math>.
 
(Solution by gundraja)
 
  
 
== 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 14:32, 5 March 2016

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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png