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Contests & Programs AMC and other contests, summer programs, etc.
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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 2025
0 replies
AIME score for college apps
Happyllamaalways   56
N 5 hours ago by Countmath1
What good colleges do I have a chance of getting into with an 11 on AIME? (Any chances for Princeton)

Also idk if this has weight but I had the highest AIME score in my school.
56 replies
Happyllamaalways
Mar 13, 2025
Countmath1
5 hours ago
MIT Beaverworks Summer Institute
PowerOfPi_09   0
5 hours ago
Hi! I was wondering if anyone here has completed this program, and if so, which track did you choose? Do rising juniors have a chance, or is it mainly rising seniors that they accept? Also, how long did it take you to complete the prerequisites?
Thanks!
0 replies
PowerOfPi_09
5 hours ago
0 replies
Convolution of order f(n)
trumpeter   70
N 5 hours ago by HamstPan38825
Source: 2019 USAMO Problem 1
Let $\mathbb{N}$ be the set of positive integers. A function $f:\mathbb{N}\to\mathbb{N}$ satisfies the equation \[\underbrace{f(f(\ldots f}_{f(n)\text{ times}}(n)\ldots))=\frac{n^2}{f(f(n))}\]for all positive integers $n$. Given this information, determine all possible values of $f(1000)$.

Proposed by Evan Chen
70 replies
trumpeter
Apr 17, 2019
HamstPan38825
5 hours ago
k HOT TAKE: MIT SHOULD NOT RELEASE THEIR DECISIONS ON PI DAY
alcumusftwgrind   8
N Today at 10:13 AM by maxamc
rant lol

Imagine a poor senior waiting for their MIT decisions just to have their hopes CRUSHED on 3/14 and they can't even celebrate pi day...

and even worse, this year's pi day is special because this year is a very special number...

8 replies
alcumusftwgrind
Today at 2:11 AM
maxamc
Today at 10:13 AM
No more topics!
System of least common multiples
va2010   24
N Nov 6, 2020 by Pleaseletmewin
Source: 2016 AMC 12A #22
How many ordered triples $(x, y, z)$ of positive integers satisfy $\text{lcm}(x, y) = 72$, $\text{lcm}(x, z)= 600$, and $\text{lcm}(y, z) = 900$?

$\textbf{(A) } 15 \qquad\textbf{(B) } 16 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 27 \qquad\textbf{(E) } 64$
24 replies
va2010
Feb 3, 2016
Pleaseletmewin
Nov 6, 2020
System of least common multiples
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Source: 2016 AMC 12A #22
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va2010
1276 posts
#1 • 2 Y
Y by Adventure10, Mango247
How many ordered triples $(x, y, z)$ of positive integers satisfy $\text{lcm}(x, y) = 72$, $\text{lcm}(x, z)= 600$, and $\text{lcm}(y, z) = 900$?

$\textbf{(A) } 15 \qquad\textbf{(B) } 16 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 27 \qquad\textbf{(E) } 64$
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Royalreter1
1913 posts
#2 • 2 Y
Y by Adventure10, Mango247
Also 2016 AMC 10A #25
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dongrae
14 posts
#3 • 1 Y
Y by Adventure10
I think #22 is (A) 15
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DrMath
2130 posts
#4 • 9 Y
Y by Not_a_Username, bestwillcui1, AstrapiGnosis, ThisIsASentence, W.Sun, DarkRunner, Bradygho, Adventure10, Mango247
Really badly misplaced for its difficulty.

Note $8\mid x$ and there are 5 ways to choose the powers of 2 that divide $y,z$ such that their maximum is 2. The answer must be divisible by 5 which automatically gives A.

If you want to check, there are 3 ways to choose the power of 3 and 1 way to choose the power of 5, confirming the answer is $5\cdot 3\cdot 1=15$ (A)
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eyzhang
150 posts
#5 • 2 Y
Y by Adventure10, Mango247
Really similar to 1987 AIME number 7!
I just recently did that prob^, so I skipped straight to this problem when I saw it :D

And yeah, I agree, badly misplaced for it's difficulty.
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Wave-Particle
3690 posts
#6 • 2 Y
Y by Not_a_Username, Adventure10
Just set up a bunch of max equations and you get 3*5=15
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bestwillcui1
2735 posts
#7 • 3 Y
Y by ACthinker, Adventure10, Mango247
yeah how is this a #22 or a #25....took 30 seconds.
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nosaj
2008 posts
#8 • 1 Y
Y by Adventure10
After spending far too much time on #25 on the 2015 AMC 10B, I decided not to look at Problem 25 at all this year. Rip :(
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Mudkipswims42
8867 posts
#9 • 1 Y
Y by Adventure10
Wait I am confused on how they got 5,3, and 1.... I got 27 by doing $2*3*6*\dfrac{3}{4}$,..
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DeathLlama9
1755 posts
#10 • 2 Y
Y by Adventure10, Mango247
BTW to make life easier just make $w$ equivalent to $\frac{z}{25}$ and you can rewrite stuff while ensuring everything's distinct
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acegikmoqsuwy2000
767 posts
#11 • 1 Y
Y by Adventure10
DeathLlama9 wrote:
BTW to make life easier just make $w$ equivalent to $\frac{z}{25}$ and you can rewrite stuff while ensuring everything's distinct

yeah i did that then did casework on the powers of two and got 5 possibilities, and then noting that 5|15 and none of the other options we get A :D
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dhwang314
1046 posts
#12 • 1 Y
Y by Adventure10
This is the problem I could have gotten right, but the problem placement confused me
EDIT:
IncompleteSolution
This post has been edited 2 times. Last edited by dhwang314, Feb 3, 2016, 6:55 PM
Reason: edit
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smo
179 posts
#14 • 1 Y
Y by Adventure10
DrMath wrote:
there are 5 ways to choose the powers of 2 that divide $y,z$ such that their maximum is 2.

I got 9 for this, could someone explain why it's 5?
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Wave-Particle
3690 posts
#15 • 2 Y
Y by Adventure10, Mango247
smo wrote:
DrMath wrote:
there are 5 ways to choose the powers of 2 that divide $y,z$ such that their maximum is 2.

I got 9 for this, could someone explain why it's 5?

$(2,2), (2,1), (1,2), (0,2), (2,0)$. Have any others to add?
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smo
179 posts
#16 • 2 Y
Y by Adventure10, Mango247
Ohh, my mistake was in thinking that both of them could be zero, so I had extras which wouldn't work. Thanks.
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618173
1751 posts
#17 • 1 Y
Y by Mango247
I just solved an AMC 10 #25 I'm so cool lol

Solution
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samrocksnature
8791 posts
#19
Y by
DrMath wrote:
Really badly misplaced for its difficulty.

Note $8\mid x$ and there are 5 ways to choose the powers of 2 that divide $y,z$ such that their maximum is 2. The answer must be divisible by 5 which automatically gives A.

If you want to check, there are 3 ways to choose the power of 3 and 1 way to choose the power of 5, confirming the answer is $5\cdot 3\cdot 1=15$ (A)

I'm not quite sure what this aopser means by "5 ways to choose the powers of 2 that divide $y,z$ such that their maximum is 2" and why the answer is divisible by $5$


EDIT: nvm I get the divisible by $5$ part, but not the other
This post has been edited 1 time. Last edited by samrocksnature, Nov 6, 2020, 6:04 AM
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Pleaseletmewin
1574 posts
#20
Y by
Alright, I'll type this up again because why not.
We have that $\text{lcm}(x,y)=2^3\cdot 3^2, \text{lcm}(x,z)=2^3\cdot 3\cdot 5^2,$ and $\text{lcm}(y,z)=2^2\cdot 3^2\cdot 5^2$.
Focus on the powers of two first. $\max\{x,y\}=2^3, \max\{x,z\}=2^3$, and $\max\{y,z\}=2^2$. It's easy to see that $x=2^3$. If not, then, one of $y$ and $z$ must be $2^3$, which is a contradiction based on $\max\{y,z\}=2^2$. Now, we just have to choose what $y$ and $z$ are. One of them must be $2^2$ and the other can be anything lower or equal to this. Simple counting shows that there are $5$ ways for this one.
Moving to the power of $3$, we see that $\max\{x,y\}=3^2, \max\{x,z\}=3$, and $\max\{y,z\}=2^2$. By the same logic, we see that $y=2^2$. It's also not hard to see that there are $3$ options for $x$ and $z$
Finally, for $5$ we see that $x,y$ have no power of $5$ and $z$ must be $5^2$ so there is only $1$ option for this case.
Hence, our answer is $3\cdot5=15$ as desired.
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Pleaseletmewin
1574 posts
#21
Y by
samrocksnature wrote:
DrMath wrote:
Really badly misplaced for its difficulty.

Note $8\mid x$ and there are 5 ways to choose the powers of 2 that divide $y,z$ such that their maximum is 2. The answer must be divisible by 5 which automatically gives A.

If you want to check, there are 3 ways to choose the power of 3 and 1 way to choose the power of 5, confirming the answer is $5\cdot 3\cdot 1=15$ (A)

I'm not quite sure what this aopser means by "5 ways to choose the powers of 2 that divide $y,z$ such that their maximum is 2" and why the answer is divisible by $5$

What do you not understand?
You know that the maximum power of $2$ of $y$ and $z$ must be $2$.
Just brute force it.
$(y,z)=(2^2,2^2), (2^2, 2^1), (2^2,2^0), (2^0,2^2),(2^1,2^2)$.
Done.
@2above, I'm pretty sure if you understand why there are 5 ways to choose this, then you understand what DrMath is saying.
This post has been edited 1 time. Last edited by Pleaseletmewin, Nov 6, 2020, 6:07 AM
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samrocksnature
8791 posts
#22 • 1 Y
Y by Mango247
Oh I see, but is there an alternative to brute forcing it?

EDIT: $2$ ways to choose the one that has $2^2$ and $0, 1, 2$ choices for the other? Doesn't that get $6$ ways?
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Pleaseletmewin
1574 posts
#23 • 1 Y
Y by Mango247
samrocksnature wrote:
Oh I see, but is there an alternative to brute forcing it?

My dude, there are literally 5 choices.
I mean, I can do this constructively for bigger numbers but my point was, it's not hard to see what the solution are.

@above, you overcounted when both of them are $2^2$.
This post has been edited 2 times. Last edited by Pleaseletmewin, Nov 6, 2020, 6:11 AM
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samrocksnature
8791 posts
#24
Y by
:cursing: Thx so much I need to study some more
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SenorIncongito
307 posts
#25
Y by
samrocksnature wrote:
Oh I see, but is there an alternative to brute forcing it?

EDIT: $2$ ways to choose the one that has $2^2$ and $0, 1, 2$ choices for the other? Doesn't that get $6$ ways?

Sometimes brute force is the only option you have... There isn't always a "clever" solution.
This post has been edited 1 time. Last edited by SenorIncongito, Nov 6, 2020, 6:21 AM
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samrocksnature
8791 posts
#26
Y by
As I can tell by your pfp

Thanks for the tip though! I'm pretty new to all this mathing

@below thx :)
This post has been edited 2 times. Last edited by samrocksnature, Nov 6, 2020, 6:20 AM
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Pleaseletmewin
1574 posts
#27
Y by
samrocksnature wrote:
As I can tell by your pfp

Thanks for the tip though! I'm pretty new to all this mathing

I mean, if you really want me to, here we go.
Choose two nonnegative integers $x$ and $y$ such that at least one of them is $n$ and the other is less than or equal to $n$.
First, discard the case when both are $n$. Then, simply, we have $2(n-1)+1=2n-1$ ways to do this.

There you go. No brute force. Satisfied?
The reason I just listed out the solutions for $y$ and $z$ was because $3$ was such a small number, it wouldn't do any harm.
This post has been edited 4 times. Last edited by Pleaseletmewin, Nov 6, 2020, 6:21 AM
Reason: i can't grammar
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