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

Revision as of 18:48, 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

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

@@ Line 21: / Line 21: @@
 ~[[User:Mathkiddus|mathkiddus]]
+==Solution 2==
+First, notice that (0,0) is a solution.
+Divide the equation by <math>y^2</math>, getting <math>12(\frac{x}{y})^2-\frac{x}{y}-6 = 0</math>. (We can ignore the <math>y=0</math> case for now.) Let <math>a = \frac{x}{y}</math>. We now have <math>12a^2-a-6=0</math>. Factoring, we get <math>(4a-3)(3a+2) = 0</math>. Therefore, the graph is satisfied when <math>4a=3</math> or <math>3a=-2</math>. Substituting <math>\frac{x}{y} = a</math> back into the equations, we get <math>4x=3y</math> or <math>3x=-2y</math>.
+Remember that both <math>x</math> and <math>y</math> are bounded by <math>-100</math> and <math>100</math>, inclusive. For <math>4x=3y</math>, the solutions are <math>(-75,-100), (-72,-96), (-69, -92), \dots, (72,96), (75,100)</math>. Remember to not count the <math>x=y=0</math> case for now. There are <math>25</math> positive solutions and <math>25</math> negative solutions for a total of <math>50</math>.
+For <math>3x-2y</math>, we do something similar. The solutions are <math>(-66,99), (-64,96), \dots, (64, -96), (66, -99)</math>. There are <math>33</math> solutions when <math>x</math> is positive and <math>33</math> solutions when <math>x</math> is negative, for a total of <math>66</math>.
+Now we can count the edge case of <math>(0,0</math>. The answer is therefore <math>50+66+1 = \boxed{117}</math>.
+~lprado

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

Revision as of 18:48, 13 February 2025

Problem

Solution 1

Solution 2