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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. 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)!

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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
1 viewing
jlacosta
Apr 2, 2025
0 replies
Modular Arithmetic Handout
MathCosine   10
N 26 minutes ago by lord_of_the_rook
Hi everyone,

I recently created a handout on modular arithmetic for a local math club. I thought it would help quite a lot with understanding basic properties, as modular arithmetic is a very popular intermediate step in number theory problems, so I decided to leave it here as a resource for anyone who needs it. Feel free to share it around, and hope it helps!

Sincerely,
MathCosine
10 replies
+2 w
MathCosine
3 hours ago
lord_of_the_rook
26 minutes ago
239 lyceum math Olympiad
R8kt   9
N 37 minutes ago by Lhaj3
Hello,
Does anyone know if it’s possible to find the solutions to the 239 Lyceum open math Olympiad somewhere online? Even if it’s in Russian, it would still be okay?
9 replies
+1 w
R8kt
Mar 9, 2023
Lhaj3
37 minutes ago
Westford Academy to host Middle School Math Competition
cyou   8
N an hour ago by Irreplaceable
Hi AOPS community,

We are excited to announce that Westford Academy (located in Westford, MA) will be hosting its first ever math competition for middle school students (grades 5-8).

Based in Massachusetts, this tournament hosts ambitious and mathematically skilled students in grades 5–8 to compete against other middle school math teams while fostering their problem-solving skills and preparing them to continue enriching their STEM skills in high school and in the future.

This competition will be held on April 12, 2025 from 12:00 PM to 5:00 PM and will feature 3 rounds (team, speed, and accuracy). The problems will be of similar difficulty for AMC 8-10 and were written by USA(J)MO and AIME qualifiers.

If you are in the Massachusetts area and are curious about Mathematics, we cordially invite you to sign up by scanning the QR code on the attached flyer. Please note that teams consist of 4-6 competitors, but if you prefer to register as an individual competitor, you will be randomly placed on a team of other individual competitors. Feel free to refer the attached flyer and website as needed.


https://sites.google.com/westfordk12.us/wamt/home?authuser=2
8 replies
cyou
Mar 25, 2025
Irreplaceable
an hour ago
[$10K+ IN PRIZES] Poolesville Math Tournament (PVMT) 2025
qwerty123456asdfgzxcvb   6
N 2 hours ago by fruitmonster97
Hi everyone!

After the resounding success of the first three years of PVMT, the Poolesville High School Math Team is excited to announce the fourth annual Poolesville High School Math Tournament (PVMT)! The PVMT team includes a MOPper and multiple USA(J)MO and AIME qualifiers!

PVMT is open to all 6th-9th graders in the country (including rising 10th graders). Students will compete in teams of up to 4 people, and each participant will take three subject tests as well as the team round. The contest is completely free, and will be held virtually on June 7, 2025, from 10:00 AM to 4:00 PM (EST).

Additionally, thanks to our sponsors, we will be awarding approximately $10K+ worth of prizes (including gift cards, Citadel merch, AoPS coupons, Wolfram licenses) to top teams and individuals. More details regarding the actual prizes will be released as we get closer to the competition date.

Further, newly for this year we might run some interesting mini-events, which we will announce closer to the competition date, such as potentially a puzzle hunt and integration bee!

If you would like to register for the competition, the registration form can be found at https://pvmt.org/register.html or https://tinyurl.com/PVMT25.

Additionally, more information about PVMT can be found at https://pvmt.org

If you have any questions not answered in the below FAQ, feel free to ask in this thread or email us at falconsdomath@gmail.com!

We look forward to your participation!

FAQ
6 replies
qwerty123456asdfgzxcvb
Apr 5, 2025
fruitmonster97
2 hours ago
No more topics!
Catch those negatives
cappucher   44
N Yesterday at 7:04 PM by Apple_maths60
Source: 2024 AMC 10A P11
How many ordered pairs of integers $(m, n)$ satisfy $\sqrt{n^2 - 49} = m$?

$
\textbf{(A) }1 \qquad
\textbf{(B) }2 \qquad
\textbf{(C) }3 \qquad
\textbf{(D) }4 \qquad
\textbf{(E) } \text{Infinitely many} \qquad
$
44 replies
cappucher
Nov 7, 2024
Apple_maths60
Yesterday at 7:04 PM
Catch those negatives
G H J
G H BBookmark kLocked kLocked NReply
Source: 2024 AMC 10A P11
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cappucher
91 posts
#1 • 1 Y
Y by IanJung
How many ordered pairs of integers $(m, n)$ satisfy $\sqrt{n^2 - 49} = m$?

$
\textbf{(A) }1 \qquad
\textbf{(B) }2 \qquad
\textbf{(C) }3 \qquad
\textbf{(D) }4 \qquad
\textbf{(E) } \text{Infinitely many} \qquad
$
This post has been edited 3 times. Last edited by cappucher, Nov 8, 2024, 4:46 AM
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mithu542
1567 posts
#2
Y by
I got $(D)$. Just squared and used difference of squares :)

I almost did 2 until I realized that negative solutions are allowed lol
This post has been edited 1 time. Last edited by mithu542, Nov 7, 2024, 5:10 PM
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Vivaandax
73 posts
#3 • 3 Y
Y by IanJung, wangzrpi, Aaronjudgeisgoat
Maa really loved to troll people with the negatives on this test
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xiaoya.peng
120 posts
#4
Y by
I got 2 because m can't be negative and the absolute value of n can't be less than 7 so that got rid of some solutions
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mithu542
1567 posts
#5 • 1 Y
Y by sadhase25
oh oops i forgot about that
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MathRook7817
643 posts
#6
Y by
mithu542 wrote:
I got $(D)$. Just squared and used difference of squares :)

I almost did 2 until I realized that negative solutions are allowed lol

bro same
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LostDreams
145 posts
#7
Y by
Square it and apply differences of squares and you notice that there are 4 possibilities when you bash which is (24, 25), (24, -25), (0, 7), and (0,-7)

Giving us the answer D
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Siddharthmaybe
106 posts
#9
Y by
Lol people really thought they tricked the question but the question tricked them, m can't be -ve cuz its square root function
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Alex-131
5311 posts
#10
Y by
I have no idea whether I did C or D, rip. I got the negative solutions, but I don't remember whether I put (0,-7) in. kms
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Siddharthmaybe
106 posts
#11
Y by
Alex-131 wrote:
I have no idea whether I did C or D, rip. I got the negative solutions, but I don't remember whether I put (0,-7) in. kms

ya i just realized that too i don't remember that :(
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stjwyl
1260 posts
#12
Y by
wait so wats the answer
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ostriches88
1527 posts
#13
Y by
stjwyl wrote:
wait so wats the answer

D
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ethanzhang1001
1060 posts
#14
Y by
Vivaandax wrote:
Maa really loved to troll people with the negatives on this test

they also should've put 8 as an answer choice in E instead because what if you accidentally set m negative :P
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ElaineGu
388 posts
#15
Y by
7 24 25 makes it trivial lmao
25, -25, 7, -7 for n
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wangzrpi
158 posts
#16
Y by
Factor by difference of squares and you see that there are 8
There are 4 extraneous
So I picked the largest one (4)
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pingpongmerrily
3540 posts
#17
Y by
i put 3 bc i thought that square roots couldn't be negative or smtg

:noo:
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golden_star_123
201 posts
#18
Y by
Nobody has put a full solution so here it is:

$\sqrt{n^2-49} = m \implies n^2 - m^2 = 49$. We have a difference of squares, so $(n-m)(n+m) = 49$.

If $(n-m)(n+m) = 7\cdot 7$, we have $n=7, n=-7$ and $m=0$. This gives two possibilities.

If $(n-m)(n+m) = 49\cdot 1$, we have $n=25, m=-24$. This doesn't work, however, as a square root is always positive.

If $(n-m)(n+m) = 1\cdot 49$, we have $n=25, m=24$. One possibility.

If $(n-m)(n+m) = -49\cdot -1$, we have $n=-25, m=24$. One possibility.

This adds up to a total of $\boxed{4}$ possibilities.
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cyberhacker
400 posts
#19
Y by
bro i put 4 and changed to 2?? i was fenting
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williamxiao
2508 posts
#20
Y by
Why did I put 3 it’s literally a square problem so it has to be even :(
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ElaineGu
388 posts
#21
Y by
williamxiao wrote:
Why did I put 3 it’s literally a square problem so it has to be even :(

:noo:
which case did u forget
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pingpongmerrily
3540 posts
#22
Y by
williamxiao wrote:
Why did I put 3 it’s literally a square problem so it has to be even :(

sameee :sob:
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ChaitraliKA
1004 posts
#23
Y by
williamxiao wrote:
Why did I put 3 it’s literally a square problem so it has to be even :(

Unless if 0 had worked

Anyways, I got this question right in the wrong way :rotfl: who cares honestly
This post has been edited 1 time. Last edited by ChaitraliKA, Nov 7, 2024, 9:34 PM
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awzhang10
72 posts
#24
Y by
i had brainrot during the test and put infinitely many LOL
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Akang11
35 posts
#25
Y by
Sillied on the negative roots, please send help :sob:
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SkatingKitty
223 posts
#26
Y by
I crashed out and I left it blank
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goldenuni678
237 posts
#27
Y by
I got D, from squaring and then using difference of squares
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SkatingKitty
223 posts
#28
Y by
I forgot the formula tears
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andrewcheng
525 posts
#29
Y by
I got C because I forgot -7,0
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gicyuraok2
1059 posts
#30
Y by
dang i probably would have sillied this one but anyway $0,\pm7$ and $\pm24,25$ fun stuff
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smarty101
322 posts
#31
Y by
forgot about 7-24-25 triangles :wallbash:
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BrighterFrog11
1116 posts
#32
Y by
aww man. I got this one wrong. I didn't think about 0 and +/- 7
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lpieleanu
2875 posts
#34
Y by
Solution
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jb2015007
1815 posts
#35
Y by
lol I actually got this right in like 2 mins cuz I was being cautious
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NS0004
186 posts
#36
Y by
How can m be negative because of the square root?
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pingpongmerrily
3540 posts
#37
Y by
NS0004 wrote:
How can m be negative because of the square root?

this is what i thought during the test but think a little harder, there are still 4 cases without m being negative
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Serengeti22
1131 posts
#38
Y by
I but 2 , I didn’t realize the negatives. The answer is 4
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Challengees24
1098 posts
#39
Y by
Did anyone else make a graph lol

n^2+m^2=49, circle with 4 roots

(0,7),(7,0),(-7,0),(0,-7)
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golden_star_123
201 posts
#40
Y by
Your approach doesn't work because m can't be negative.
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Challengees24
1098 posts
#41
Y by
oml lmho i did it wrong way and still got the answer write

daym
This post has been edited 1 time. Last edited by Challengees24, Nov 9, 2024, 5:33 PM
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Challengees24
1098 posts
#42 • 1 Y
Y by PikaPika999
I just realized it should have been $n^2-m^2=49$

heyyy maybe this test wasnt so bad lol(joke kinda sorta, cuz it wasnt that bad)
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Apple_maths60
16 posts
#43 • 1 Y
Y by PikaPika999
I don't think (7,0)or (-7,0) would be a solution because if we put zero in the main equation we get √(-49)=m×m .
But √(-49) is not possible
So the answer must be 2
(25,-24)&(25,24)
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NS0004
186 posts
#44 • 1 Y
Y by PikaPika999
Bruh amc said the answer was 4....
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AbhayAttarde01
1477 posts
#45 • 1 Y
Y by PikaPika999
Apple_maths60 wrote:
I don't think (7,0)or (-7,0) would be a solution because if we put zero in the main equation we get √(-49)=m×m .
But √(-49) is not possible
So the answer must be 2
(25,-24)&(25,24)

swap the numbers
think you accidently swapper the numbers for m and n
it shoult be (0, -7), (0, 7), (24, -25), (24, 25)
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sadas123
1198 posts
#46 • 1 Y
Y by PikaPika999
I got (4) possible values
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Apple_maths60
16 posts
#47 • 1 Y
Y by PikaPika999
AbhayAttarde01 wrote:
Apple_maths60 wrote:
I don't think (7,0)or (-7,0) would be a solution because if we put zero in the main equation we get √(-49)=m×m .
But √(-49) is not possible
So the answer must be 2
(25,-24)&(25,24)

swap the numbers
think you accidently swapper the numbers for m and n
it shoult be (0, -7), (0, 7), (24, -25), (24, 25)


Oh, ok got it. I actually did that mistake
Thank you!
This post has been edited 1 time. Last edited by Apple_maths60, Yesterday at 7:06 PM
Reason: .
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