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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Tuesday at 2:14 PM
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Tuesday at 2:14 PM
0 replies
nth root of a, b in O(sqrt(n))
tenniskidperson3   63
N Yesterday at 9:29 PM by ihategeo_1969
Source: 2014 USAMO problem 6
Prove that there is a constant $c>0$ with the following property: If $a, b, n$ are positive integers such that $\gcd(a+i, b+j)>1$ for all $i, j\in\{0, 1, \ldots n\}$, then\[\min\{a, b\}>c^n\cdot n^{\frac{n}{2}}.\]
63 replies
tenniskidperson3
Apr 30, 2014
ihategeo_1969
Yesterday at 9:29 PM
Sad Combinatorics II
tastymath75025   26
N Yesterday at 9:13 PM by ray66
Source: 2018 USAMO 6, by Richard Stong
Let $a_n$ be the number of permutations $(x_1, x_2, \dots, x_n)$ of the numbers $(1,2,\dots, n)$ such that the $n$ ratios $\frac{x_k}{k}$ for $1\le k\le n$ are all distinct. Prove that $a_n$ is odd for all $n\ge 1$.

Proposed by Richard Stong
26 replies
tastymath75025
Apr 19, 2018
ray66
Yesterday at 9:13 PM
Sad Combinatorics
62861   109
N Yesterday at 8:56 PM by ray66
Source: USAMO 2018 P4 and JMO 2018 P5, by Ankan Bhattacharya
Let $p$ be a prime, and let $a_1, \dots, a_p$ be integers. Show that there exists an integer $k$ such that the numbers
\[a_1 + k, a_2 + 2k, \dots, a_p + pk\]produce at least $\tfrac{1}{2} p$ distinct remainders upon division by $p$.

Proposed by Ankan Bhattacharya
109 replies
62861
Apr 19, 2018
ray66
Yesterday at 8:56 PM
Goals for 2025-2026
Airbus320-214   287
N Yesterday at 8:21 PM by ouia1234
Please write down your goal/goals for competitions here for 2025-2026.
287 replies
Airbus320-214
May 11, 2025
ouia1234
Yesterday at 8:21 PM
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
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Mathandski
775 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
1381 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
651 posts
#3 • 1 Y
Y by cubres
I got r=1. Does this mean I’ve made it!?
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MathRook7817
766 posts
#4 • 2 Y
Y by Pengu14, cubres
haha u cant fool me
i got the url basically memorized
Attachments:
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williamxiao
2516 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
651 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
2516 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
651 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!!!!!
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Mathandski
775 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
766 posts
#10 • 1 Y
Y by cubres
lets go i made mop!
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sadas123
1381 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
1215 posts
#12 • 1 Y
Y by cubres
i made MOP!
woo hoo!
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Mathandski
775 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
1030 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
205 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
1446 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
1446 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
2028 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
1446 posts
#21 • 1 Y
Y by cubres
Same here :| my r is 0 and my x is 0 :stretcher:
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Yrock
1446 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
775 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
1446 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
190 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
Attachments:
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Mathandski
775 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
1061 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
1446 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
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#30 • 2 Y
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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
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#31
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Why did I know it was a rickroll, and do it anyway.
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