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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
[TEST RELEASED] OMMC Year 5
DottedCaculator   65
N 38 minutes ago by Math-lover1
Test portal: https://ommc-test-portal-2025.vercel.app/

Hello to all creative problem solvers,

Do you want to work on a fun, untimed team math competition with amazing questions by MOPpers and IMO & EGMO medalists? $\phantom{You lost the game.}$
Do you want to have a chance to win thousands in cash and raffle prizes (no matter your skill level)?

Check out the fifth annual iteration of the

Online Monmouth Math Competition!

Online Monmouth Math Competition, or OMMC, is a 501c3 accredited nonprofit organization managed by adults, college students, and high schoolers which aims to give talented high school and middle school students an exciting way to develop their skills in mathematics.

Our website: https://www.ommcofficial.org/
Our Discord (6000+ members): https://tinyurl.com/joinommc

This is not a local competition; any student 18 or younger anywhere in the world can attend. We have changed some elements of our contest format, so read carefully and thoroughly. Join our Discord or monitor this thread for updates and test releases.

How hard is it?

We plan to raffle out a TON of prizes over all competitors regardless of performance. So just submit: a few minutes of your time will give you a great chance to win amazing prizes!

How are the problems?

You can check out our past problems and sample problems here:
https://www.ommcofficial.org/sample
https://www.ommcofficial.org/2022-documents
https://www.ommcofficial.org/2023-documents
https://www.ommcofficial.org/ommc-amc

How will the test be held?/How do I sign up?

Solo teams?

Test Policy

Timeline:
Main Round: May 17th - May 24th
Test Portal Released. The Main Round of the contest is held. The Main Round consists of 25 questions that each have a numerical answer. Teams will have the entire time interval to work on the questions. They can submit any time during the interval. Teams are free to edit their submissions before the period ends, even after they submit.

Final Round: May 26th - May 28th
The top placing teams will qualify for this invitational round (5-10 questions). The final round consists of 5-10 proof questions. Teams again will have the entire time interval to work on these questions and can submit their proofs any time during this interval. Teams are free to edit their submissions before the period ends, even after they submit.

Conclusion of Competition: Early June
Solutions will be released, winners announced, and prizes sent out to winners.

Scoring:

Prizes:

I have more questions. Whom do I ask?

We hope for your participation, and good luck!

OMMC staff

OMMC’S 2025 EVENTS ARE SPONSORED BY:

[list]
[*]Nontrivial Fellowship
[*]Citadel
[*]SPARC
[*]Jane Street
[*]And counting!
[/list]
65 replies
+2 w
DottedCaculator
Apr 26, 2025
Math-lover1
38 minutes ago
Jane street swag package? USA(J)MO
arfekete   36
N 3 hours ago by dolphinday
Hey! People are starting to get their swag packages from Jane Street for qualifying for USA(J)MO, and after some initial discussion on what we got, people are getting different things. Out of curiosity, I was wondering how they decide who gets what.
Please enter the following info:

- USAMO or USAJMO
- Grade
- Score
- Award/Medal/HM
- MOP (yes or no, if yes then color)
- List of items you got in your package

I will reply with my info as an example.
36 replies
arfekete
May 7, 2025
dolphinday
3 hours ago
[CASH PRIZES] IndyINTEGIRLS Spring Math Competition
Indy_Integirls   15
N 4 hours ago by GallopingUnicorn45
[center]IMAGE

Greetings, AoPS! IndyINTEGIRLS will be hosting a virtual math competition on May 25,
2024 from 12 PM to 3 PM EST.
Join other woman-identifying and/or non-binary "STEMinists" in solving problems, socializing, playing games, winning prizes, and more! If you are interested in competing, please register here![/center]

----------

[center]Important Information[/center]

Eligibility: This competition is open to all woman-identifying and non-binary students in middle and high school. Non-Indiana residents and international students are welcome as well!

Format: There will be a middle school and high school division. In each separate division, there will be an individual round and a team round, where students are grouped into teams of 3-4 and collaboratively solve a set of difficult problems. There will also be a buzzer/countdown/Kahoot-style round, where students from both divisions are grouped together to compete in a MATHCOUNTS-style countdown round! There will be prizes for the top competitors in each division.

Problem Difficulty: Our amazing team of problem writers is working hard to ensure that there will be problems for problem-solvers of all levels! The middle school problems will range from MATHCOUNTS school round to AMC 10 level, while the high school problems will be for more advanced problem-solvers. The team round problems will cover various difficulty levels and are meant to be more difficult, while the countdown/buzzer/Kahoot round questions will be similar to MATHCOUNTS state to MATHCOUNTS Nationals countdown round in difficulty.

Platform: This contest will be held virtually through Zoom. All competitors are required to have their cameras turned on at all times unless they have a reason for otherwise. Proctors and volunteers will be monitoring students at all times to prevent cheating and to create a fair environment for all students.

Prizes: At this moment, prizes are TBD, and more information will be provided and attached to this post as the competition date approaches. Rest assured, IndyINTEGIRLS has historically given out very generous cash prizes, and we intend on maintaining this generosity into our Spring Competition.

Contact & Connect With Us: Follow us on Instagram @indy.integirls, join our Discord, follow us on TikTok @indy.integirls, and email us at indy@integirls.org.

---------
[center]Help Us Out

Please help us in sharing the news of this competition! Our amazing team of officers has worked very hard to provide this educational opportunity to as many students as possible, and we would appreciate it if you could help us spread the word!
15 replies
Indy_Integirls
May 11, 2025
GallopingUnicorn45
4 hours ago
prime spam
fruitmonster97   36
N Today at 2:44 PM by ZMB038
Source: 2024 AMC 10A #3
What is the sum of the digits of the smallest prime that can be written as a sum of $5$ distinct primes?

$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$
36 replies
fruitmonster97
Nov 7, 2024
ZMB038
Today at 2:44 PM
The Riemann Zeta Function
aoum   2
N May 2, 2025 by aoum
The Riemann Zeta Function: A Central Object in Mathematics

The Riemann Zeta Function $\zeta(s)$ is one of the most important functions in mathematics, deeply connected to number theory, complex analysis, and mathematical physics. Its study has led to profound insights into the distribution of prime numbers and the structure of the complex plane.

1. Definition

For complex numbers $s$ with real part greater than $1$, the Riemann zeta function is defined by the absolutely convergent series:

$$
\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}.
$$
That is,

$$
\zeta(s) = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \cdots
$$
This definition converges when $\Re(s) > 1$.

2. Analytic Continuation

The function $\zeta(s)$ can be extended to a meromorphic function on the entire complex plane, except for a simple pole at $s=1$. The extension is achieved using techniques like:

[list]
[*] The functional equation,
[*] Mellin transforms,
[*] Dirichlet series manipulations.
[/list]

3. Functional Equation

The Riemann zeta function satisfies a remarkable symmetry, given by the functional equation:

$$
\zeta(s) = 2^s \pi^{s-1} \sin\left( \frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s),
$$
where $\Gamma(s)$ is the Gamma function.

This equation connects the values of $\zeta(s)$ at $s$ and $1-s$, and it is crucial for the study of its zeros.

4. Euler Product Formula

One of Euler's most important discoveries was that $\zeta(s)$ has an infinite product representation over prime numbers when $\Re(s) > 1$:

$$
\zeta(s) = \prod_{p \, \text{prime}} \frac{1}{1 - p^{-s}}.
$$
This shows the deep connection between $\zeta(s)$ and the distribution of prime numbers. It expresses the fundamental theorem of arithmetic (unique prime factorization) analytically.

5. Special Values

At positive even integers:

$$
\zeta(2) = \frac{\pi^2}{6},
\quad
\zeta(4) = \frac{\pi^4}{90},
\quad
\zeta(6) = \frac{\pi^6}{945},
\quad \text{etc.}
$$
At negative integers:

$$
\zeta(-n) = -\frac{B_{n+1}}{n+1},
$$
where $B_n$ are the Bernoulli numbers.

For example:

$$
\zeta(-1) = -\frac{1}{12},
\quad
\zeta(-3) = \frac{1}{120}.
$$
Note that $\zeta(0) = -\frac{1}{2}$.

6. Zeros of the Zeta Function

The zeros of $\zeta(s)$ are of two types:

[list]
[*] Trivial zeros: Located at negative even integers $s = -2, -4, -6, \dots$.
[*] Non-trivial zeros: Located in the "critical strip" where $0 < \Re(s) < 1$.
[/list]

The famous Riemann Hypothesis conjectures that all non-trivial zeros lie on the "critical line" $\Re(s) = \frac{1}{2}$.

7. Applications of $\zeta(s)$

The Riemann zeta function appears in:

[list]
[*] Prime number theory: The distribution of primes.
[*] Random matrix theory: Models of quantum chaos.
[*] Physics: Statistical mechanics and quantum field theory.
[*] Probability: Connections to branching processes and the zeta distribution.
[*] Fractal geometry: Dimension computations involve zeta-like functions.
[/list]

8. Proof Sketch: $\zeta(2) = \frac{\pi^2}{6}$

One classic proof involves expanding $\sin(\pi x)$ as an infinite product:

$$
\sin(\pi x) = \pi x \prod_{n=1}^\infty \left( 1 - \frac{x^2}{n^2} \right),
$$
Taking the logarithm and differentiating, and then comparing coefficients, yields:

$$
\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.
$$
Thus:

$$
\zeta(2) = \frac{\pi^2}{6}.
$$
9. References

[list]
[*] Wikipedia: Riemann Zeta Function
[*] H. M. Edwards, Riemann's Zeta Function
[*] Titchmarsh, The Theory of the Riemann Zeta-Function
[*] AoPS Wiki: Riemann Zeta Function
[/list]
2 replies
aoum
Apr 26, 2025
aoum
May 2, 2025
No more topics!
do NOT double count (0,0)
bobthegod78   40
N Apr 20, 2025 by NicoN9
Source: 2025 AIME I P4
Find the number of ordered pairs $(x,y)$, where both $x$ and $y$ are integers between $-100$ and $100$ inclusive, such that $12x^2-xy-6y^2=0$.
40 replies
bobthegod78
Feb 7, 2025
NicoN9
Apr 20, 2025
do NOT double count (0,0)
G H J
G H BBookmark kLocked kLocked NReply
Source: 2025 AIME I P4
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bobthegod78
2982 posts
#1 • 1 Y
Y by cubres
Find the number of ordered pairs $(x,y)$, where both $x$ and $y$ are integers between $-100$ and $100$ inclusive, such that $12x^2-xy-6y^2=0$.
This post has been edited 1 time. Last edited by bobthegod78, Feb 7, 2025, 3:53 PM
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MathPerson12321
3786 posts
#2 • 1 Y
Y by cubres
Solution
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mathMagicOPS
850 posts
#3 • 1 Y
Y by cubres
got 118 rip
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razormouth
81 posts
#4 • 1 Y
Y by cubres
Solving for x in terms of y using the quadratic formula gives x = -2y/3 or x = 3y/4, then +-(2,-3) , (4,-6), ... (66,-99) and +- (3,4), (6,8), ....(75,100) and (0,0) for a total of 117
Z K Y
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Elephant200
1472 posts
#5 • 1 Y
Y by cubres
mathMagicOPS wrote:
got 118 rip

Same... I double counted (0,0)
Z K Y
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fryingpan546
361 posts
#6 • 1 Y
Y by cubres
Solution
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sixoneeight
1138 posts
#7 • 1 Y
Y by cubres
118 gang
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cappucher
96 posts
#8 • 1 Y
Y by cubres
Took me way too long to factor the expression...

$12x^2 - xy - 6y^2$ can be factored as $(4x - 3y)(3x + 2y)$. Thus, we consider three cases: $4x - 3y = 0$, $3x + 2y = 0$, or both.

The first case yields $25 \cdot 2 + 1$ pairs. The second case yields $33 \cdot 2 + 1$ pairs. The third case yields $1$ pair. This yields $51 + 67 - 1 = \boxed{117}$ pairs.
This post has been edited 1 time. Last edited by cappucher, Feb 7, 2025, 8:17 PM
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Countmath1
180 posts
#9 • 1 Y
Y by cubres
First note that $x = y = 0$ is a solution then divide by $xy$, sub $a = \frac{x}{y}$ to get $12a - 1 -\frac{6}{a} = 0$, so $\frac{x}{y} = -\frac{2}{3}, \frac{3}{4}$. Casework gives $33\cdot 2 + 25\cdot 2 + 1 = \boxed{\textbf{(117)}}$.
Z K Y
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xHypotenuse
783 posts
#10 • 1 Y
Y by cubres
the moment I saw the title I wanted to shoot myself

i doubled counted (0,0)....rip
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Mathandski
766 posts
#11 • 16 Y
Y by Leo.Euler, MathRook7817, OronSH, KnowingAnt, megahertz13, zhoujef000, Sedro, Lhaj3, aidan0626, Alex-131, vrondoS, anduran, tricky.math.spider.gold.1, megarnie, cubres, vincentwant
I spent 4 minutes looking into Vieta jumping and Pell's before I realized this was the AIME whoops
Attachments:
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andrewcheng
525 posts
#12 • 1 Y
Y by cubres
when you forget that you don't need to double count when x is a multiple of 6
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MathPerson12321
3786 posts
#13 • 1 Y
Y by cubres
Just consider non-zero y... its not that hard and then add $1$ at the end
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williamxiao
2517 posts
#14 • 1 Y
Y by cubres
Put 118 :/
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JTmath07
39 posts
#15 • 1 Y
Y by cubres
andrewcheng wrote:
when you forget that you don't need to double count when x is a multiple of 6

fr
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MathRook7817
744 posts
#16 • 1 Y
Y by cubres
almost put 118
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maromex
197 posts
#17 • 1 Y
Y by cubres
im part of 118 club :skull:

I guess from now on I have to consider PIE for every single problem I do casework on
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lpieleanu
3001 posts
#18 • 1 Y
Y by cubres
Solution
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junlongsun
70 posts
#19 • 1 Y
Y by cubres
$$12x^2-xy-6y=0$$$$(4x-3y)(3y+2x)=0$$$$3y=4x, y=\frac{4}{3}x$$and $$3x=-2y, y=-\frac{3}{2}x$$
For first case, $y=\frac{4}{3}x$

$y$ can equal ${-100, -96, -92,..., 96, 100}$

Which is 51 cases

For the second case, $y=-\frac{3}{2}$

$y$ can equal ${-99, -96, ..., 96, 99}$

Which is 67 cases

$$51 + 67 = 118$$
But the lines intersect at $(0,0)$ so we have to subtract $1$ for overcount.

$$118 - 1 = 117$$
$$\fbox{117}$$
This post has been edited 3 times. Last edited by junlongsun, Feb 7, 2025, 8:05 PM
Reason: Edit
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MathPerson12321
3786 posts
#20 • 1 Y
Y by cubres
MathPerson12321 wrote:
Solution

yeah this avoids that completely
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remedy
19 posts
#21 • 1 Y
Y by cubres
really easy problem imo
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Elephant200
1472 posts
#22 • 1 Y
Y by cubres
It was definitely a straightforward problem; it's unfortunate so many of us put 117
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BS2012
1046 posts
#23 • 1 Y
Y by cubres
isn't the answer 117 though
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mathMagicOPS
850 posts
#24 • 1 Y
Y by cubres
BS2012 wrote:
isn't the answer 117 though

unfortunate that so many people got it correct???!?! :rotfl:
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MathPerson12321
3786 posts
#25 • 1 Y
Y by cubres
Elephant200 wrote:
It was definitely a straightforward problem; it's unfortunate so many of us put 117

:skull:
it is 117
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sanaops9
835 posts
#26 • 1 Y
Y by cubres
bro i put 116, missed (0, 0) case oops.
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Mathkiddie
322 posts
#27 • 1 Y
Y by cubres
mathMagicOPS wrote:
got 118 rip
same here! I can't believe I double counted (0, 0) :blush:
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tzliu
163 posts
#28 • 1 Y
Y by cubres
Elephant200 wrote:
mathMagicOPS wrote:
got 118 rip

Same... I double counted (0,0)

same
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apple143
62 posts
#29 • 1 Y
Y by cubres
Elephant200 wrote:
It was definitely a straightforward problem; it's unfortunate so many of us put 117

it is 117 lol. i got it as well.
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akliu
1801 posts
#30 • 1 Y
Y by cubres
I have fortunately been traumatized by combinatorics problems in various mocks a lot; enough that I immediately didn't trust 118.
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pog
4917 posts
#31 • 1 Y
Y by cubres
i may not have double counted (0, 0) but i also did not single count (0, 0)
This post has been edited 1 time. Last edited by pog, Feb 8, 2025, 6:23 PM
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Yrock
1294 posts
#32 • 2 Y
Y by maromex, cubres
we could make a poll and see how much people put 118

I almost put 116 (thought (0,0) gave undefined)
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Existing_Human1
214 posts
#33 • 2 Y
Y by MathPerson12321, cubres
We have $$12x^2 - xy - 6y^2 = 0$$,assume some constant $k$ allows us to factor by grouping, thus the expression becomes $$12x^2 + kxy - (k+1)xy - 6y^2 = 0$$or $x(12x+ky) -y((k+1)x + 6y)$. In order to factor we want $$ \frac{k}{12} = \frac{6}{k+1}$$,we find $k = 8$ and we can factor and solve as above
This post has been edited 2 times. Last edited by Existing_Human1, Feb 9, 2025, 7:43 PM
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sadas123
1309 posts
#34 • 1 Y
Y by cubres
I just did some rigorous bashing for factoring and got 118 but then.. I changed it to 117 :)
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eg4334
637 posts
#35 • 1 Y
Y by cubres
basically like (4x-3y)(3x+2y)=0 so like 4x=3y or 3x=-2y ykyk and count each case seperately like for the first x magnitude goes up to 75 but is a multiple of 3 so like 51 and then the second one is like x magnitude go up to 66 but even so like 67 so like 51+67-1 cuz 0, 0 is overcounted so like $\boxed{117}$
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sadas123
1309 posts
#36 • 1 Y
Y by cubres
here is my solution I didn't actually bash that hard tho Solution + Answer
This post has been edited 1 time. Last edited by sadas123, Feb 9, 2025, 10:05 PM
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fruitmonster97
2498 posts
#37 • 1 Y
Y by cubres
QF in terms of $x.$ We get $x=-\tfrac23y$ or $x=\tfrac34y,$ so $50+66+1=\boxed{117}.$

of course in test i did 25+33+1 because i forgot y could be negative. anyone else get 059?
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MathPerson12321
3786 posts
#38 • 1 Y
Y by cubres
Existing_Human1 wrote:
We have $$12x^2 - xy - 6y^2 = 0$$,assume some constant $k$ allows us to factor by grouping, thus the expression becomes $$12x^2 + kxy - (k+1)xy - 6y^2 = 0$$or $x(12x+ky) -y((k+1)x + 6y)$. In order to factor we want $$ \frac{k}{12} = \frac{6}{k+1}$$,we find $k = 8$ and we can factor and solve as above

Best solution trust
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Juno_34
83 posts
#39 • 1 Y
Y by cubres
factor to get $\left(4x-3y\right)\left(3x+2y\right)=0$ then just find x and y and make sure not to double counted the zero :wallbash:
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jasperE3
11352 posts
#40
Y by
bobthegod78 wrote:
Find the number of ordered pairs $(x,y)$, where both $x$ and $y$ are integers between $-100$ and $100$ inclusive, such that $12x^2-xy-6y^2=0$.

By the quadratic formula (viewing this equation as a quadratic in $y$) this equation is equivalent to:
$$6y^2+xy-12x^2=0\Leftrightarrow y=\frac{-x\pm17x}{12}\Leftrightarrow y\in\left\{\frac{4x}3,-\frac{3x}2\right\}.$$We now have two mutually exclusive cases:

Case 1: $y=\frac{4x}3$
We must have $3\mid x$. Since $y\le100$, we can constraint $\frac{4x}3\le100$ which rearranges to $x\le75$. Likewise, since $y\ge-100$, $x\ge-75$. All such $-75\le x\le75$ with $3\mid x$ will produce a valid and unique $-100\le y\le100$ satisfying the original equation, so it just remains to count the number of solutions for $x$ which is $\frac{75-(-75)}3+1=51$.

Case 2: $y\ne\frac{4x}3$ and $y=\frac{-3x}2$
As before $2\mid x$ and we proceed to bound $x$. Since $-\frac{3x}2=y\le100$, we have $x\ge-66$, and similarly $x\le66$. All such $-66\le x\le66$ with $2\mid x$ will produce a valid and unique solution for $y$ except for $x=0$, which violates the $y\ne\frac{4x}3$ constraint. Thus our answer for this case is (subtracting $1$ at the end so as not to count $x=0$) $\frac{66-(-66)}2+1-1=66$.

In all there are $51+66=\boxed{117}$ solutions.
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NicoN9
157 posts
#41
Y by
Fix $y$, and solve over $x$, we have\[
x=\frac{-y\pm17y}{24}
\]so either $x=\frac{3}{4}y$, or $x=-\frac{2}{3}y$.

$\bullet$ if $x=\frac{3}{4}y$, then we have $y=-100, -96, \dots 100$, hence there are $51$ solutions.

$\bullet$ Similarly for $x=-\frac{2}{3}y$, there are $67$ solutions.

We only double counted $(0, 0)$ as a solution, so the answer is $51+67-1=117$.
This post has been edited 1 time. Last edited by NicoN9, Apr 21, 2025, 5:36 AM
Reason: accidentally swapped x and y
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