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Difference between revisions of "2025 AIME I Problems/Problem 4"

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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>
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~[[User:Mathkiddus|mathkiddus]]

Revision as of 18:10, 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 $67+51-1=\boxed{117}$

~mathkiddus

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