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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.

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0 replies
jlacosta
Apr 2, 2025
0 replies
Next Steps for AIME
GeniusGiraffe   8
N 5 minutes ago by Aopsauser9999
Hi everyone,

I am a freshmen who got a 181 index this year, with a 111 on AMC 10 and a 7 on AIME.

Now that I have around 6 months to prepare before AMC 10, what should I do to prepare? I am pretty sure I can get >130 on AMC 10 right now.
My goal is to make USA(J)MO next year and do decent on it.
8 replies
GeniusGiraffe
Today at 5:23 AM
Aopsauser9999
5 minutes ago
Discuss the Stanford Math Tournament Here
Aaronjudgeisgoat   253
N 9 minutes ago by mathprodigy2011
I believe discussion is allowed after yesterday at midnight, correct?
If so, I will put tentative answers on this thread.
By the way, does anyone know the answer to Geometry Problem 5? I was wondering if I got that one right
Also, if you put answers, please put it in a hide tag

Answers for the Algebra Subject Test
Estimated Algebra Cutoffs
Answers for the Geometry Subject Test
Estimated Geo Cutoffs
Answers for the Discrete Subject Test
Estimated Cutoffs for Discrete
Answers for the Team Round
Guts Answers
253 replies
Aaronjudgeisgoat
Apr 14, 2025
mathprodigy2011
9 minutes ago
MathPath
PatTheKing806   4
N 3 hours ago by evt917
Is anybody else going to MathPath?

I haven't gotten in. its been 3+ weeks since they said my application was done.
4 replies
PatTheKing806
Mar 24, 2025
evt917
3 hours ago
Combinatorics
AlexCenteno2007   0
4 hours ago
In how many ways can 8 white rooks be placed on an 8x8 chessboard such that the main diagonal of the board is not occupied?
0 replies
AlexCenteno2007
4 hours ago
0 replies
NYMC High School Summer Program (HSSP)
missionsqhc   0
5 hours ago
I was recently accepted into the black level of New York Math Circle's High School Summer Program (HSSP). I would be taking this program as a prefrosh and wondering if anyone who has experience with this program could comment on it. It's my understanding that the program will heavily focus on problem solving and competitions rather than math research. In that case, would it offer any benefit to an incoming college math major?
0 replies
missionsqhc
5 hours ago
0 replies
Range of function
girishpimoli   3
N 5 hours ago by rchokler
Range of function $\displaystyle f(x)=\frac{e^{2x}-e^{x}+1}{e^{2x}+e^{x}+1}$
3 replies
girishpimoli
Today at 11:51 AM
rchokler
5 hours ago
Solve an equation
lgx57   2
N Today at 2:56 PM by lgx57
Find all positive integers $n$ and $x$ such that:
$$2^{2n+1}-7=x^2$$
2 replies
lgx57
Mar 12, 2025
lgx57
Today at 2:56 PM
Indonesia Regional MO 2019 Part A
parmenides51   17
N Today at 2:42 PM by Rohit-2006
Indonesia Regional MO
Year 2019 Part A

Time: 90 minutes Rules


p1. In the bag there are $7$ red balls and $8$ white balls. Audi took two balls at once from inside the bag. The chance of taking two balls of the same color is ...


p2. Given a regular hexagon with a side length of $1$ unit. The area of the hexagon is ...


p3. It is known that $r, s$ and $1$ are the roots of the cubic equation $x^3 - 2x + c = 0$. The value of $(r-s)^2$ is ...


p4. The number of pairs of natural numbers $(m, n)$ so that $GCD(n,m) = 2$ and $LCM(m,n) = 1000$ is ...


p5. A data with four real numbers $2n-4$, $2n-6$, $n^2-8$, $3n^2-6$ has an average of $0$ and a median of $9/2$. The largest number of such data is ...


p6. Suppose $a, b, c, d$ are integers greater than $2019$ which are four consecutive quarters of an arithmetic row with $a <b <c <d$. If $a$ and $d$ are squares of two consecutive natural numbers, then the smallest value of $c-b$ is ...


p7. Given a triangle $ABC$, with $AB = 6$, $AC = 8$ and $BC = 10$. The points $D$ and $E$ lies on the line segment $BC$. with $BD = 2$ and $CE = 4$. The measure of the angle $\angle DAE$ is ...


p8. Sequqnce of real numbers $a_1,a_2,a_3,...$ meet $\frac{na_1+(n-1)a_2+...+2a_{n-1}+a_n}{n^2}=1$ for each natural number $n$. The value of $a_1a_2a_3...a_{2019}$ is ....


p9. The number of ways to select four numbers from $\{1,2,3, ..., 15\}$ provided that the difference of any two numbers at least $3$ is ...


p10. Pairs of natural numbers $(m , n)$ which satisfies $$m^2n+mn^2 +m^2+2mn = 2018m + 2019n + 2019$$are as many as ...


p11. Given a triangle $ABC$ with $\angle ABC =135^o$ and $BC> AB$. Point $D$ lies on the side $BC$ so that $AB=CD$. Suppose $F$ is a point on the side extension $AB$ so that $DF$ is perpendicular to $AB$. The point $E$ lies on the ray $DF$ such that $DE> DF$ and $\angle ACE = 45^o$. The large angle $\angle AEC$ is ...


p12. The set of $S$ consists of $n$ integers with the following properties: For every three different members of $S$ there are two of them whose sum is a member of $S$. The largest value of $n$ is ....


p13. The minimum value of $\frac{a^2+2b^2+\sqrt2}{\sqrt{ab}}$ with $a, b$ positive reals is ....


p14. The polynomial P satisfies the equation $P (x^2) = x^{2019} (x+ 1) P (x)$ with $P (1/2)= -1$ is ....


p15. Look at a chessboard measuring $19 \times 19$ square units. Two plots are said to be neighbors if they both have one side in common. Initially, there are a total of $k$ coins on the chessboard where each coin is only loaded exactly on one square and each square can contain coins or blanks. At each turn. You must select exactly one plot that holds the minimum number of coins in the number of neighbors of the plot and then you must give exactly one coin to each neighbor of the selected plot. The game ends if you are no longer able to select squares with the intended conditions. The smallest number of $k$ so that the game never ends for any initial square selection is ....
17 replies
parmenides51
Nov 11, 2021
Rohit-2006
Today at 2:42 PM
How to prove one-one function
Vulch   6
N Today at 2:20 PM by Vulch
Hello everyone,
I am learning functional equations.
To prove the below problem one -one function,I have taken two non-negative real numbers $ (1,2)$ from the domain $\Bbb R_{*},$ and put those numbers into the given function f(x)=1/x.It gives us 1=1/2.But it's not true.So ,it can't be one-one function.But in the answer,it is one-one function.Would anyone enlighten me where is my fault? Thank you!
6 replies
Vulch
Apr 11, 2025
Vulch
Today at 2:20 PM
Inequalities
sqing   6
N Today at 2:20 PM by sqing
Let $ a,b,c> 0 $ and $  \frac{a}{a^2+ab+c}+\frac{b}{b^2+bc+a}+\frac{c}{c^2+ca+b} \geq 1$. Prove that
$$  a+b+c\leq 3    $$
6 replies
sqing
Apr 4, 2025
sqing
Today at 2:20 PM
hard number theory
eric201291   0
Today at 2:17 PM
Prove:There are no integers x, y, that y^2+9998587980=x^3.
0 replies
eric201291
Today at 2:17 PM
0 replies
Amc 10 mock
Mathsboy100   3
N Today at 1:50 PM by iwastedmyusername
let \[\lfloor  x   \rfloor\]denote the greatest integer less than or equal to x . What is the sum of the squares of the real numbers x for which \[  x^2 - 20\lfloor x \rfloor + 19 = 0  \]
3 replies
Mathsboy100
Oct 9, 2024
iwastedmyusername
Today at 1:50 PM
Inequalities
lgx57   4
N Today at 1:45 PM by pooh123
Let $0 < a,b,c < 1$. Prove that

$$a(1-b)+b(1-c)+c(1-a)<1$$
4 replies
lgx57
Mar 19, 2025
pooh123
Today at 1:45 PM
Let x,y,z be non-zero reals
Purple_Planet   3
N Today at 1:08 PM by sqing
Let $x,y,z$ be non-zero real numbers. Define $E=\frac{|x+y|}{|x|+|y|}+\frac{|x+z|}{|x|+|z|}+\frac{|y+z|}{|y|+|z|}$, then the number of all integers which lies in the range of $E$ is equal to.
3 replies
Purple_Planet
Jul 16, 2019
sqing
Today at 1:08 PM
Catch those negatives
cappucher   44
N Apr 6, 2025 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
Apr 6, 2025
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
1568 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
79 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
1568 posts
#5 • 1 Y
Y by sadhase25
oh oops i forgot about that
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MathRook7817
653 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
5332 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
1263 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
3565 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
203 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
3565 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
2902 posts
#34
Y by
Solution
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jb2015007
1862 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
3565 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
1135 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
203 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
24 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
1483 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
1214 posts
#46 • 1 Y
Y by PikaPika999
I got (4) possible values
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Apple_maths60
24 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, Apr 6, 2025, 7:06 PM
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