Difference between revisions of "2016 AIME I Problems/Problem 7"

Revision as of 17:44, 4 March 2016

Problem

For integers $a$ and $b$ consider the complex number $\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i$

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: $ab > -2016$ In this case, if $0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = {\frac{\sqrt{|a+b|}}{ab+100}}$ 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 $ab = -a^2 \ne -100$ ,

@@ Line 4: / Line 4: @@
 Find the number of ordered pairs of integers <math>(a,b)</math> such that this complex number is a real number.
+== Solution == We consider two cases:
+Case 1:  <math>ab > -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>
+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>,
 == See also ==
 {{AIME box|year=2016|n=I|num-b=6|num-a=8}}
 {{MAA Notice}}

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Difference between revisions of "2016 AIME I Problems/Problem 7"

Revision as of 17:44, 4 March 2016

Problem

See also