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>
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<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.
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We consider two cases:
 
We consider two cases:
  
Case 1:  <math>ab \ge -2016</math>   
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'''Case 1:''' <math>ab \ge -2016</math>.  
  
 
In this case, if
 
In this case, if
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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 < -2016</math>.  
  
 
In this case, we want
 
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>
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<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  
 
Squaring, we have the equations <math>ab \ne -100</math> (which always holds in this case) and  
<cmath>-(ab + 2016) = |a + b|.</cmath>
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<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>,  
 
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>
 
<cmath>ac - 2016 = c - a \Rightarrow (a - 1)(c + 1) = 2015 \Rightarrow (a,b) = (2, -2014), (6, -402), (14, -154), (32, -64). </cmath>
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Thus, the answer is <math>87 + 16 = \boxed{103}</math>.
 
Thus, the answer is <math>87 + 16 = \boxed{103}</math>.
  
Solution by gundraja
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(Solution by gundraja)
 +
 
 +
==Solution 2==
 +
Similar to Solution 1, but concise:
 +
 
 +
First, we set the imaginary expression to <math>0</math>, so that <math>|a+b|=0</math> or <math>a=-b</math>, of which there are <math>44\cdot 2+1=89</math> possibilities. But <math>(a,b)\ne(\pm 10,\mp 10)</math> because the denominator would be <math>0</math>. So this gives <math>89-2=87</math> solutions.
 +
 
 +
Then we try to cancel the imaginary part with the square root of the real part, which must be negative. So <math>ab<-2016</math>.
 +
<math>ab+2016=-|a+b|\rightarrow ab\pm a\pm b+2016=0\rightarrow (a\pm 1)(b\pm 1)=-2015</math> by Simon's Favorite Factoring Trick.
 +
 
 +
We must have the negative part lesser in magnitude than the positive part, because an increase in magnitude of a lesser number is MORE than a decrease in the magnitude of a positive number, so the product will net to be more magnitude, namely <math>ab<-2016</math> and <math>-2015\approx -2016</math>.
 +
 
 +
The factors of <math>(\text{-})2015</math> are <math>(\text{-})5\cdot 13\cdot 31</math>, so the <math>(a+1,b+1)=\{(-1,2015),(-5,403), (-13,155),(-31,65)\}</math> and the sets flipped.
 +
 
 +
Similarly from the second case of <math>(a-1,b-1)</math> we also have <math>8</math> solutions.
 +
 
 +
Thus, <math>(a,b),(b,a)=\{(\mp 2,\pm 2014),(\mp 6,\pm 402),(\mp 14,\pm 154),(\mp 32,\pm 64)\}</math>. Surely, all of their products, <math>ab=-4028,-2412,-2156,-2048<-2016</math>.
 +
 
 +
So there are <math>87+16=\boxed{103}</math> solutions.
  
 
== 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 00:22, 15 February 2017

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 $87 + 16 = \boxed{103}$.

(Solution by gundraja)

Solution 2

Similar to Solution 1, but concise:

First, we set the imaginary expression to $0$, so that $|a+b|=0$ or $a=-b$, of which there are $44\cdot 2+1=89$ possibilities. But $(a,b)\ne(\pm 10,\mp 10)$ because the denominator would be $0$. So this gives $89-2=87$ solutions.

Then we try to cancel the imaginary part with the square root of the real part, which must be negative. So $ab<-2016$. $ab+2016=-|a+b|\rightarrow ab\pm a\pm b+2016=0\rightarrow (a\pm 1)(b\pm 1)=-2015$ by Simon's Favorite Factoring Trick.

We must have the negative part lesser in magnitude than the positive part, because an increase in magnitude of a lesser number is MORE than a decrease in the magnitude of a positive number, so the product will net to be more magnitude, namely $ab<-2016$ and $-2015\approx -2016$.

The factors of $(\text{-})2015$ are $(\text{-})5\cdot 13\cdot 31$, so the $(a+1,b+1)=\{(-1,2015),(-5,403), (-13,155),(-31,65)\}$ and the sets flipped.

Similarly from the second case of $(a-1,b-1)$ we also have $8$ solutions.

Thus, $(a,b),(b,a)=\{(\mp 2,\pm 2014),(\mp 6,\pm 402),(\mp 14,\pm 154),(\mp 32,\pm 64)\}$. Surely, all of their products, $ab=-4028,-2412,-2156,-2048<-2016$.

So there are $87+16=\boxed{103}$ solutions.

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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