Difference between revisions of "2016 AIME I Problems/Problem 7"
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We consider two cases: | We consider two cases: | ||
| − | Case 1: <math>ab \ge -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>, 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 \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{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 | ||
<cmath>-(ab + 2016) = |a + b|.</cmath> | <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.</cmath> | + | <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}} | ||
Revision as of 18:30, 4 March 2016
Problem
For integers 
and 
consider the complex number
![\[\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i\]](http://latex.artofproblemsolving.com/9/4/a/94aed5cb77ab692adad418e0f67918ea8fe39dc6.png)
Find the number of ordered pairs of integers 
such that this complex number is a real number.
Solution
We consider two cases:
Case 1: 
In this case, if
![\[0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = -\frac{\sqrt{|a+b|}}{ab+100}\]](http://latex.artofproblemsolving.com/e/4/b/e4bf248d3b281431410bee79a2982a80ace108fc.png)
then 
and 
. Thus 
so 
. Thus 
, yielding 
values. However since 
, we have 
. Thus there are 
allowed tuples in this case.
Case 2: 
.
In this case, we want
![\[0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = \frac{\sqrt{ab+2016} - \sqrt{|a+b|}}{ab+100}\]](http://latex.artofproblemsolving.com/5/3/f/53f4a8cf50d2f90504dd42364c70e37fd7a3629a.png)
Squaring, we have the equations (which always holds in this case) and
![\[-(ab + 2016) = |a + b|.\]](http://latex.artofproblemsolving.com/d/3/d/d3db3ff88c0db1b27f35220e45e3eb86284b975e.png)
Then if 
and 
, let 
. If 
,
![\[ac - 2016 = c - a \Rightarrow (a - 1)(c + 1) = 2015 \Rightarrow (a,b) = (2, -2014), (6, -402), (14, -154), (32, -64).\]](http://latex.artofproblemsolving.com/7/1/8/718187e91fc7f1b29feaedb3c779189c45d51d46.png)
Note that 
for every one of these solutions. If 
, then
![\[ac - 2016 = a - c \Rightarrow (a + 1)(c - 1) = 2015 \Rightarrow (a,b) = (2014, -2), (402, -6), (154, -14), (64, -32).\]](http://latex.artofproblemsolving.com/b/e/b/bebd0f411c6b1612a26a9f12a120300423a4ef55.png)
Again, for every one of the above solutions. This yields 
solutions. Similarly, if 
and 
, there are solutions. Thus, there are a total of 
solutions in this case.
Thus, the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
