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Difference between revisions of "2025 AIME I Problems/Problem 4"
(→‎Solution 1)
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We begin by factoring, <math>12x^2-xy-6y^2=(3x+2y)(4x-3y)=0.</math> Since the RHS is <math>0</math> we have two options,
 
We begin by factoring, <math>12x^2-xy-6y^2=(3x+2y)(4x-3y)=0.</math> Since the RHS is <math>0</math> we have two options,
 
   
 
   
−
Case 1: <math>3x+2y = 0</math>
+
<math>\underline{\text{Case 1:}}\text{ } 3x+2y = 0</math>
 +
 
 
In this case we have, <math>y=\frac{-3x}{2}.</math> Using the bounding on <math>y</math> we have,  
 
In this case we have, <math>y=\frac{-3x}{2}.</math> Using the bounding on <math>y</math> we have,  
 
<cmath>-100\le\frac{-3x}{2}\le 100.</cmath>  
 
<cmath>-100\le\frac{-3x}{2}\le 100.</cmath>  
 
<cmath>\frac{200}{3}\ge x \ge \frac{-200}{3}.</cmath>
 
<cmath>\frac{200}{3}\ge x \ge \frac{-200}{3}.</cmath>
−
In addition in order for <math>y</math> to be integer <math>2|x,</math> so we substitute <math>x=2k.</math> <cmath>\frac{200}{3}\ge 2k \ge \frac{-200}{3}.</cmath> <cmath>\frac{100}{3}\ge k \ge \frac{-100}{3}.</cmath> From this we have solutions starting from <math>-33</math> to <math>33</math> which is <math>67</math> solutions.
+
In addition in order for <math>y</math> to be integer <math>2 | x,</math> so we substitute <math>x=2k.</math> <cmath>\frac{200}{3}\ge 2k \ge \frac{-200}{3}.</cmath> <cmath>\frac{100}{3}\ge k \ge \frac{-100}{3}.</cmath> From this we have solutions starting from <math>-33</math> to <math>33</math> which is <math>67</math> solutions.
 +
 
 +
<math>\underline{\text{Case 2: }}\text{ } 4x-3y = 0</math>
  
−
Case 2: <math>4x-3y = 0</math>
 
 
On the other hand, we have, <math>y=\frac{4x}{3}.</math> From bounds we have,  
 
On the other hand, we have, <math>y=\frac{4x}{3}.</math> From bounds we have,  
 
<cmath>-100\le\frac{4x}{3}\le 100.</cmath>  
 
<cmath>-100\le\frac{4x}{3}\le 100.</cmath>  
 
<cmath>-75 \le x \le 75.</cmath>
 
<cmath>-75 \le x \le 75.</cmath>
−
In this case, for <math>y</math> to be integer <math>3|x,</math> so we substitute <math>x=3t.</math> <cmath>-75 \le 3t \le 75.</cmath> <cmath>-25 \le t \le 25.</cmath> This gives us <math>51</math> solutions.
+
In this case, for <math>y</math> to be integer <math>3 | x,</math> so we substitute <math>x=3t.</math> <cmath>-75 \le 3t \le 75.</cmath> <cmath>-25 \le t \le 25.</cmath> This gives us <math>51</math> solutions.
  
 
Finally we overcount one case which is the intersection of the <math>2</math> lines or the point <math>(0,0).</math> Therefore our answer is <math>67+51-1=\boxed{117}</math>
 
Finally we overcount one case which is the intersection of the <math>2</math> lines or the point <math>(0,0).</math> Therefore our answer is <math>67+51-1=\boxed{117}</math>

Revision as of 18:09, 13 February 2025

Problem

Find the number of ordered pairs $(x,y)$, where both $x$ and $y$ are integers between $-100$ and $100$ inclusive, such that $12x^2-xy-6y^2=0$.

Solution 1

We begin by factoring, $12x^2-xy-6y^2=(3x+2y)(4x-3y)=0.$ Since the RHS is $0$ we have two options,

$\underline{\text{Case 1:}}\text{ } 3x+2y = 0$

In this case we have, $y=\frac{-3x}{2}.$ Using the bounding on $y$ we have, \[-100\le\frac{-3x}{2}\le 100.\] \[\frac{200}{3}\ge x \ge \frac{-200}{3}.\] In addition in order for $y$ to be integer $2 | x,$ so we substitute $x=2k.$ \[\frac{200}{3}\ge 2k \ge \frac{-200}{3}.\] \[\frac{100}{3}\ge k \ge \frac{-100}{3}.\] From this we have solutions starting from $-33$ to $33$ which is $67$ solutions.

$\underline{\text{Case 2: }}\text{ } 4x-3y = 0$

On the other hand, we have, $y=\frac{4x}{3}.$ From bounds we have, \[-100\le\frac{4x}{3}\le 100.\] \[-75 \le x \le 75.\] In this case, for $y$ to be integer $3 | x,$ so we substitute $x=3t.$ \[-75 \le 3t \le 75.\] \[-25 \le t \le 25.\] This gives us $51$ solutions.

Finally we overcount one case which is the intersection of the $2$ lines or the point $(0,0).$ Therefore our answer is $67+51-1=\boxed{117}$

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