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== | + | ==Solution== |
− | Case 1: <math>ab \ge -2016</math> In this case, if | + | We consider two cases: |
− | <cmath>0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = \frac{\sqrt{|a+b|}}{ab+100 | + | |
+ | '''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. | 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{|a+b|}}{ab+100}}</ | + | |
+ | 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>. | ||
+ | |||
+ | (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
Contents
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 .
(Solution by gundraja)
Solution 2
Similar to Solution 1, but concise:
First, we set the imaginary expression to , so that or , of which there are possibilities. But because the denominator would be . So this gives solutions.
Then we try to cancel the imaginary part with the square root of the real part, which must be negative. So . 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 and .
The factors of are , so the and the sets flipped.
Similarly from the second case of we also have solutions.
Thus, . Surely, all of their products, .
So there are solutions.
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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