2025 AIME I Problems/Problem 4

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

Solution 2

First, notice that (0,0) is a solution.

Divide the equation by $y^2$ , getting $12(\frac{x}{y})^2-\frac{x}{y}-6 = 0$ . (We can ignore the $y=0$ case for now.) Let $a = \frac{x}{y}$ . We now have $12a^2-a-6=0$ . Factoring, we get $(4a-3)(3a+2) = 0$ . Therefore, the graph is satisfied when $4a=3$ or $3a=-2$ . Substituting $\frac{x}{y} = a$ back into the equations, we get $4x=3y$ or $3x=-2y$ .

Remember that both $x$ and $y$ are bounded by $-100$ and $100$ , inclusive. For $4x=3y$ , the solutions are $(-75,-100), (-72,-96), (-69, -92), \dots, (72,96), (75,100)$ . Remember to not count the $x=y=0$ case for now. There are $25$ positive solutions and $25$ negative solutions for a total of $50$ .

For $3x-2y$ , we do something similar. The solutions are $(-66,99), (-64,96), \dots, (64, -96), (66, -99)$ . There are $33$ solutions when $x$ is positive and $33$ solutions when $x$ is negative, for a total of $66$ .

Now we can count the edge case of $(0,0)$ . The answer is therefore $50+66+1 = \boxed{117}$ .

~lprado

Solution 3

You can use the quadratic formula for this equation: 12x^2 - xy - 6y^2 = 0; Although this solution may seem to be misleading, it works!

$You get: (-b +- sqrt (b^2-4ac)) / 2a

= xy +- sqrt(x^2y^2+(12*6*4*x^2*y^2) / 24x^2

= (xy +- sqrt289x^2 y^2 )/24x^2

= 18xy/24x^2, and -16xy/24x^2$ (Error compiling LaTeX. Unknown error_msg)

Rather than putting this equation as zero, the numerators and denominators must be equal. These two equations simplify to:

3y = 4x; -2y = 3x;

As x and y are between -100 and 100, for the first equation, x can be between (-75,75), but x must be a multiple of 3, so there are:

((75+75)/3) + 1 = 51 solutions for this case.

For -2y = 3x:

x can be between (-66, 66), but x has to be a multiple of 2.

Therefore, there are (66+66)/2 + 1 = 67 solutions for this case

However, the one overlap would be x = 0, because y would be 0 in both solutions.

Therefore, the answer is $51+67-1 = \boxed{117}$ .

-U-King3.14Root

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=J-0BapU4Yuk

2025 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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