Difference between revisions of "2016 AIME I Problems/Problem 7"
Fclvbfm934 (talk | contribs) (→Solution) |
Mathkiddie (talk | contribs) |
||
(16 intermediate revisions by 9 users not shown) | |||
Line 1: | Line 1: | ||
==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. | ||
Line 9: | Line 9: | ||
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>. |
− | <cmath>0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = -\frac{\sqrt{|a+b|}}{ab+100}</cmath> | + | |
+ | 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. | 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 | + | '''Case 2:''' <math>ab < -2016</math>. |
− | <cmath>0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = \frac{\sqrt{ab | + | |
+ | 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 | 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>, | ||
+ | <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 18:00, 2 January 2024
Problem
For integers and
consider the complex number
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
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
Squaring, we have the equations
(which always holds in this case) and
Then if
and
, let
. If
,
Note that
for every one of these solutions. If
, then
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.