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> | + | '''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 < -2016</math>. | + | '''Case 2:''' <math>ab < -2016</math>. |
In this case, we want | In this case, we want | ||
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Thus, the answer is <math>87 + 16 = \boxed{103}</math>. | Thus, the answer is <math>87 + 16 = \boxed{103}</math>. | ||
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== 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 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.