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

Latest revision as of 17:00, 2 January 2024

Problem

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

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: $ab \ge -2016$ .

In this case, if $0 = \text{Im}\left({\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i}\right) = -\frac{\sqrt{|a+b|}}{ab+100}$ 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 $ab = -a^2 \ne -100$ , we have $a \ne \pm 10$ . Thus there are $87$ allowed tuples $(a,b)$ in this case.

Case 2: $ab < -2016$ .

In this case, we want $0 = \text{Im}\left({\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i}\right) = \frac{\sqrt{-ab-2016} - \sqrt{|a+b|}}{ab+100}$ Squaring, we have the equations $ab \ne -100$ (which always holds in this case) and $-(ab + 2016)= |a + b|.$ Then if $a > 0$ and $b < 0$ , let $c = -b$ . If $c > a$ , $ac - 2016 = c - a \Rightarrow (a - 1)(c + 1) = 2015 \Rightarrow (a,b) = (2, -2014), (6, -402), (14, -154), (32, -64).$ Note that $ab < -2016$ for every one of these solutions. If $c < a$ , then $ac - 2016 = a - c \Rightarrow (a + 1)(c - 1) = 2015 \Rightarrow (a,b) = (2014, -2), (402, -6), (154, -14), (64, -32).$ Again, $ab < -2016$ for every one of the above solutions. This yields $8$ solutions. Similarly, if $a < 0$ and $b > 0$ , there are $8$ solutions. Thus, there are a total of $16$ solutions in this case.

Thus, the answer is $87 + 16 = \boxed{103}$ .

@@ Line 28: / Line 28: @@
 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 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 ==
 {{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"

Latest revision as of 17:00, 2 January 2024

Problem

Solution

See also