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k a June Highlights and 2025 AoPS Online Class Information
jlacosta   0
Jun 2, 2025
Congratulations to all the mathletes who competed at National MATHCOUNTS! If you missed the exciting Countdown Round, you can watch the video at this link. Are you interested in training for MATHCOUNTS or AMC 10 contests? How would you like to train for these math competitions in half the time? We have accelerated sections which meet twice per week instead of once starting on July 8th (7:30pm ET). These sections fill quickly so enroll today!

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0 replies
jlacosta
Jun 2, 2025
0 replies
Nerfed 2023 JMO 1
brainfertilzer   30
N 6 minutes ago by Kempu33334
Source: 2024 AIME II P11
Find the number of triples of nonnegative integers $(a,b,c)$ satisfying $a + b + c = 300$ and
\[ a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6{,}000{,}000.\]
30 replies
1 viewing
brainfertilzer
Feb 8, 2024
Kempu33334
6 minutes ago
Aime ii - 2006/15
tiredepartment   43
N 7 minutes ago by Kempu33334
Given that $x$, $y$, and $z$ are real numbers that satisfy:

\[ x=\sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}} \]
\[ y=\sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}} \]
\[ z=\sqrt{x^2-\frac{1}{36}}+\sqrt{y^2-\frac{1}{36}} \]

and that $x+y+z=\frac{m}{\sqrt{n}}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime, find $m+n$.
43 replies
tiredepartment
Mar 27, 2006
Kempu33334
7 minutes ago
Degree Six Polynomial's Roots
ksun48   44
N 25 minutes ago by Kempu33334
Source: 2014 AIME I Problem 14
Let $m$ be the largest real solution to the equation \[\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}= x^2-11x-4.\] There are positive integers $a,b,c$ such that $m = a + \sqrt{b+\sqrt{c}}$. Find $a+b+c$.
44 replies
+1 w
ksun48
Mar 14, 2014
Kempu33334
25 minutes ago
AMC 10 Registry
Crimzion   7
N an hour ago by OGMATH
Just wondering how do i apply for amc 10 this year, maa website says info about last years.
7 replies
Crimzion
Yesterday at 3:55 AM
OGMATH
an hour ago
No more topics!
MOP Cutoffs Out?
Mathandski   30
N May 21, 2025 by ZMB038
MAA has just emailed a press release announcing the formula they will be using this year to come up with the MOP cutoff that applies to you! Here's the process:

1. Multiply your age by $1434$, let $n$ be the result.

2. Calculate $\varphi(n)$, where $\varphi$ is the Euler's totient theorem, which calculates the number of integers less than $n$ relatively prime to $n$.

3. Multiply your result by $1434$ again because why not, let the result be $m$.

4. Define the Fibonacci sequence $F_0 = 1, F_1 = 1, F_n = F_{n-1} + F_{n-2}$ for $n \ge 2$. Let $r$ be the remainder $F_m$ leaves when you divide it by $69$.

5. Let $x$ be your predicted USA(J)MO score.

6. You will be invited if your score is at least $\lfloor \frac{x + \sqrt[r]{r^2} + r \ln(r)}{r} \rfloor$.

7. Note that there may be additional age restrictions for non-high schoolers.

See here for MAA's original news message.

.

.

.


Edit (4/2/2025): This was an April Fool's post.
Here's the punchline
30 replies
Mathandski
Apr 1, 2025
ZMB038
May 21, 2025
MOP Cutoffs Out?
G H J
G H BBookmark kLocked kLocked NReply
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Mathandski
774 posts
#1 • 24 Y
Y by Andyluo, zhoujef000, williamxiao, scannose, bjump, aidan0626, ARWonder, Liontiger, Alex-131, jkim0656, Yrock, krithikrokcs, justJen, megarnie, OronSH, arfekete, Toinfinity, vincentwant, mdk2013, cubres, RollingPanda4616, mrtheory, EpicBird08, Soupboy0
MAA has just emailed a press release announcing the formula they will be using this year to come up with the MOP cutoff that applies to you! Here's the process:

1. Multiply your age by $1434$, let $n$ be the result.

2. Calculate $\varphi(n)$, where $\varphi$ is the Euler's totient theorem, which calculates the number of integers less than $n$ relatively prime to $n$.

3. Multiply your result by $1434$ again because why not, let the result be $m$.

4. Define the Fibonacci sequence $F_0 = 1, F_1 = 1, F_n = F_{n-1} + F_{n-2}$ for $n \ge 2$. Let $r$ be the remainder $F_m$ leaves when you divide it by $69$.

5. Let $x$ be your predicted USA(J)MO score.

6. You will be invited if your score is at least $\lfloor \frac{x + \sqrt[r]{r^2} + r \ln(r)}{r} \rfloor$.

7. Note that there may be additional age restrictions for non-high schoolers.

See here for MAA's original news message.

.

.

.


Edit (4/2/2025): This was an April Fool's post.
Here's the punchline
This post has been edited 4 times. Last edited by Mathandski, Apr 2, 2025, 4:38 PM
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sadas123
1333 posts
#2 • 1 Y
Y by cubres
Imagine knowing that the video was a Rick roll and still pressing it :skull
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Pengu14
644 posts
#3 • 1 Y
Y by cubres
I got r=1. Does this mean I’ve made it!?
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MathRook7817
758 posts
#4 • 2 Y
Y by Pengu14, cubres
haha u cant fool me
i got the url basically memorized
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williamxiao
2517 posts
#5 • 1 Y
Y by cubres
MathRook7817 wrote:
haha u cant fool me
i got the url basically memorized

whoa why does the color change even in the screenshot

Is it even possible for $x < \lfloor \frac{x + \sqrt[r]{r^2} + r \ln(r)}{r} \rfloor$
If we ignore the floor, for all r>1, $\sqrt[r]{r^2} \le r$ so $\frac{\sqrt[r]{r^2}}{r}$ contributes <1, $\frac{x}{r} < x-1$, and $\frac{r \ln(r)}{r} = \ln(r)$ grows very slowly compared to $x-\frac{x}{r}$

Conclusion: We all made MOP!!!!
This post has been edited 1 time. Last edited by williamxiao, Apr 1, 2025, 11:48 PM
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Pengu14
644 posts
#6 • 1 Y
Y by cubres
It’s possible for our score to be significantly lower than prediction though
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williamxiao
2517 posts
#7 • 1 Y
Y by cubres
Pengu14 wrote:
It’s possible for our score to be significantly lower than prediction though

oh that's true

Well i have a predicted score of 0 because i didn't qual so... guaranteed mop!!!
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Pengu14
644 posts
#8 • 1 Y
Y by cubres
williamxiao wrote:
Pengu14 wrote:
It’s possible for our score to be significantly lower than prediction though

oh that's true

Well i have a predicted score of 0 because i didn't qual so... guaranteed mop!!!

No dividing by zero!!!!!
Z K Y
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Mathandski
774 posts
#9 • 1 Y
Y by cubres
williamxiao wrote:
MathRook7817 wrote:
haha u cant fool me
i got the url basically memorized

whoa why does the color change even in the screenshot

Is it even possible for $x < \lfloor \frac{x + \sqrt[r]{r^2} + r \ln(r)}{r} \rfloor$
If we ignore the floor, for all r>1, $\sqrt[r]{r^2} \le r$ so $\frac{\sqrt[r]{r^2}}{r}$ contributes <1, $\frac{x}{r} < x-1$, and $\frac{r \ln(r)}{r} = \ln(r)$ grows very slowly compared to $x-\frac{x}{r}$

Conclusion: We all made MOP!!!!

As a hint, $r = 1$ happens quite a lot (but not always). As a math problem, feel free to try figuring out the possible ages where $r \neq 1$
This post has been edited 1 time. Last edited by Mathandski, Apr 2, 2025, 12:18 AM
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MathRook7817
758 posts
#10 • 1 Y
Y by cubres
lets go i made mop!
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sadas123
1333 posts
#11 • 1 Y
Y by cubres
MathRook7817 wrote:
lets go i made mop!

Me too! Except I didn't make USAJMO :)
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jkim0656
1081 posts
#12 • 1 Y
Y by cubres
i made MOP!
woo hoo!
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Mathandski
774 posts
#13 • 1 Y
Y by cubres
rbo how are so many people making MOP did I set it up wrong wth

@below I'll let you know tmrw
This post has been edited 1 time. Last edited by Mathandski, Apr 2, 2025, 1:42 AM
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scannose
1020 posts
#14 • 1 Y
Y by cubres
Mathandski wrote:
rbo how are so many people making MOP did I set it up wrong wth

is r supposed to be equal to 1
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Richard-Stillhard
9 posts
#15 • 1 Y
Y by cubres
Hello, in step 2 it asks for us to find phi(n), but there are infinitely many integers less than n relatively prime to it... so isn't it infinite? I'm confused.
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lord_of_the_rook
182 posts
#16 • 1 Y
Y by cubres
Richard-Stillhard wrote:
Hello, in step 2 it asks for us to find phi(n), but there are infinitely many integers less than n relatively prime to it... so isn't it infinite? I'm confused.

I think they made a mistake, it means positive integers.
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Richard-Stillhard
9 posts
#17 • 2 Y
Y by lord_of_the_rook, cubres
Yeah I gathered lol, I was being sarcastic.
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Yrock
1345 posts
#18 • 1 Y
Y by cubres
I need help calculating, what's the cycle of the fibonacci numbers modulo 69?
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Yrock
1345 posts
#19 • 1 Y
Y by cubres
I hope I know how to add :|

Anyone verify? Hope I can calculate my score soon!

(also I lost the game)
This post has been edited 3 times. Last edited by Yrock, Apr 2, 2025, 2:19 AM
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Amkan2022
2030 posts
#20 • 1 Y
Y by cubres
Looks like my cutoff is... Undefined?

Click to reveal hidden text
This post has been edited 1 time. Last edited by Amkan2022, Apr 2, 2025, 2:21 AM
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Yrock
1345 posts
#21 • 1 Y
Y by cubres
Same here :| my r is 0 and my x is 0 :stretcher:
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Yrock
1345 posts
#22 • 2 Y
Y by Mathandski, cubres
Hope this is the right code! If you are lazy run this..

  1. import math
  2. def totient(n):
  3. if n == 1:
  4. return 1
  5.  
  6. phi = n
  7. p = 2
  8. while p * p <= n:
  9. if n % p == 0:
  10. while n % p == 0:
  11. n //= p
  12. phi -= phi // p
  13. p += 1
  14.  
  15. if n > 1:
  16. phi -= phi // n
  17.  
  18. return phi
  19.  
  20. fibMod69 = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 20, 6, 26, 32, 58, 21, 10, 31, 41, 3, 44, 47, 22, 0, 22, 22, 44, 66, 41, 38, 10, 48, 58, 37, 26, 63, 20, 14, 34, 48, 13, 61, 5, 66, 2, 68, 1, 0]
  21. age = int(input("What is your age?"))
  22. jmoscore = int(input("What is your predicted AMO/JMO score?"))
  23. if age > 20 or age < 12:
  24. print("Sorry, are not eligible for the MOP at this age.")
  25. elif jmoscore > 42 or jmoscore < 0:
  26. print("That is not a valid JMO score.")
  27. else:
  28. n = 1434 * age
  29. p = totient(n)
  30. m = 1434 * p
  31. nr = m%48
  32. r = fibMod69[nr]
  33. final = math.floor((jmoscore+r**(2/r)+r*math.log(r))/(r))
  34. print(str(final)+" compared to "+str(jmoscore))
  35. if final<=jmoscore:
  36. print("You made MOP!!!!")
  37. else:
  38. print("Womp womp, you failed...")

Is it just me or can no one qual MOP with this
This post has been edited 4 times. Last edited by Yrock, Apr 2, 2025, 3:04 AM
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Mathandski
774 posts
#23 • 1 Y
Y by cubres
Most of the code should be correct but I believe line 30 should say
print(final <= jmoscore)
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Yrock
1345 posts
#24 • 1 Y
Y by cubres
Mathandski wrote:
Most of the code should be correct but I believe line 30 should say
print(final <= jmoscore)

oof edited

Time to change to code so that people don't enter incorrect information :|

EDIT:

is final always one more than jmoscore?!
This post has been edited 2 times. Last edited by Yrock, Apr 2, 2025, 3:05 AM
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smbellanki
187 posts
#25 • 1 Y
Y by cubres
Yrock wrote:
Mathandski wrote:
Most of the code should be correct but I believe line 30 should say
print(final <= jmoscore)

oof edited

Time to change to code so that people don't enter incorrect information :|

EDIT:

is final always one more than jmoscore?!

Yeah
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Mathandski
774 posts
#26 • 11 Y
Y by fake123, cj13609517288, bjump, OronSH, Sedro, blueprimes, Yrock, EpicBird08, cubres, Pengu14, vincentwant
In case anyone was curious, here's the joke: according to the rules, everybody missed MOP by 1 point!

We claim that $r = 1$ if and only if your age cannot be written as $3^a 239^b$. In other words, unless if you are $1, 3, 9, 27, 81, 239, 243, 717, 729, \dots$ years old, $r = 1$.

The proof comes down to proving two these parts:
Period of F_n modulo 69 is 48

m is a multiple of 48

Unless if you are $1, 3, 9, 27, 81, 239, 243, \dots$ years old, $48 \mid m$ meaning $F_m \equiv F_0 \equiv 1 \pmod{69}$. With $r = 1$ proved, we plug it in to see the cutoff is $x + 1$, which is exactly one more than your USA(J)MO score!
This post has been edited 3 times. Last edited by Mathandski, Apr 2, 2025, 5:24 PM
Reason: = -> \equiv
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BS2012
1058 posts
#27 • 1 Y
Y by cubres
Mathandski wrote:
In case anyone was curious, here's the joke: according to the rules, everybody missed MOP by 1 point!

We claim that $r = 1$ if and only if your age cannot be written as $3^a 239^b$. In other words, unless if you are $1, 3, 9, 27, 81, 239, 243, 717, 729, \dots$ years old, $r = 1$.

The proof comes down to proving two these parts:
Period of F_n modulo 69 is 48

m is a multiple of 48

Unless if you are $1, 3, 9, 27, 81, 239, 243, \dots$ years old, $48 \mid m$ meaning $F_m \equiv F_0 \equiv 1 \pmod{69}$. With $r = 1$ proved, we plug it in to see the cutoff is $x + 1$, which is exactly one more than your USA(J)MO score!

what if you are 9 years old, or predict way low
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anticodon
202 posts
#28 • 1 Y
Y by cubres
Mathandski wrote:
1. Multiply your age by $1434$, let $n$ be the result.

that's when I realized April fools
MAA doesn't age discriminate like that (even if they did, why such a big factor of 1434)
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Yrock
1345 posts
#29 • 2 Y
Y by cubres, Sedro
BS2012 wrote:
Mathandski wrote:
In case anyone was curious, here's the joke: according to the rules, everybody missed MOP by 1 point!

We claim that $r = 1$ if and only if your age cannot be written as $3^a 239^b$. In other words, unless if you are $1, 3, 9, 27, 81, 239, 243, 717, 729, \dots$ years old, $r = 1$.

The proof comes down to proving two these parts:
Period of F_n modulo 69 is 48

m is a multiple of 48

Unless if you are $1, 3, 9, 27, 81, 239, 243, \dots$ years old, $48 \mid m$ meaning $F_m \equiv F_0 \equiv 1 \pmod{69}$. With $r = 1$ proved, we plug it in to see the cutoff is $x + 1$, which is exactly one more than your USA(J)MO score!

what if you are 9 years old, or predict way low

I'm nine so I made MOP :P
This post has been edited 1 time. Last edited by Yrock, Apr 2, 2025, 10:38 PM
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Mathandski
774 posts
#30 • 2 Y
Y by Yrock, cubres
BS2012 wrote:
what if you are 9 years old

It's why I added rule 7 :D
Quote:
7. Note that there may be additional age restrictions for non-high schoolers.
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ZMB038
315 posts
#31
Y by
Why did I know it was a rickroll, and do it anyway.
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