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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. 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)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
9 How young were you and what score did you get when starting AMC 8?
TalentedElephant41   3
N 33 minutes ago by TalentedElephant41
Just curious!
3 replies
TalentedElephant41
an hour ago
TalentedElephant41
33 minutes ago
Lots of Cyclic Quads
Vfire   104
N Today at 5:53 AM by Ilikeminecraft
Source: 2018 USAMO #5
In convex cyclic quadrilateral $ABCD$, we know that lines $AC$ and $BD$ intersect at $E$, lines $AB$ and $CD$ intersect at $F$, and lines $BC$ and $DA$ intersect at $G$. Suppose that the circumcircle of $\triangle ABE$ intersects line $CB$ at $B$ and $P$, and the circumcircle of $\triangle ADE$ intersects line $CD$ at $D$ and $Q$, where $C,B,P,G$ and $C,Q,D,F$ are collinear in that order. Prove that if lines $FP$ and $GQ$ intersect at $M$, then $\angle MAC = 90^\circ$.

Proposed by Kada Williams
104 replies
Vfire
Apr 19, 2018
Ilikeminecraft
Today at 5:53 AM
Goals for 2025-2026
Airbus320-214   107
N Today at 5:02 AM by Jaxman8
Please write down your goal/goals for competitions here for 2025-2026.
107 replies
Airbus320-214
May 11, 2025
Jaxman8
Today at 5:02 AM
Evan's mean blackboard game
hwl0304   72
N Today at 3:26 AM by HamstPan38825
Source: 2019 USAMO Problem 5, 2019 USAJMO Problem 6
Two rational numbers \(\tfrac{m}{n}\) and \(\tfrac{n}{m}\) are written on a blackboard, where \(m\) and \(n\) are relatively prime positive integers. At any point, Evan may pick two of the numbers \(x\) and \(y\) written on the board and write either their arithmetic mean \(\tfrac{x+y}{2}\) or their harmonic mean \(\tfrac{2xy}{x+y}\) on the board as well. Find all pairs \((m,n)\) such that Evan can write 1 on the board in finitely many steps.

Proposed by Yannick Yao
72 replies
hwl0304
Apr 18, 2019
HamstPan38825
Today at 3:26 AM
No more topics!
sum to 2024
pog   25
N Apr 4, 2025 by Apple_maths60
Source: 2024 AMC 10A #4 / 2024 AMC 12A #3
The number $2024$ is written as the sum of not necessarily distinct two-digit numbers. What is the least number of two-digit numbers needed to write this sum?

$\textbf{(A) }20\qquad\textbf{(B) }21\qquad\textbf{(C) }22\qquad\textbf{(D) }23\qquad\textbf{(E) }24$
25 replies
pog
Nov 7, 2024
Apple_maths60
Apr 4, 2025
sum to 2024
G H J
Source: 2024 AMC 10A #4 / 2024 AMC 12A #3
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pog
4917 posts
#1
Y by
The number $2024$ is written as the sum of not necessarily distinct two-digit numbers. What is the least number of two-digit numbers needed to write this sum?

$\textbf{(A) }20\qquad\textbf{(B) }21\qquad\textbf{(C) }22\qquad\textbf{(D) }23\qquad\textbf{(E) }24$
This post has been edited 1 time. Last edited by jlacosta, Nov 7, 2024, 5:08 PM
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ConfidentKoala4
611 posts
#2
Y by
B i think iirc
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pog
4917 posts
#3 • 1 Y
Y by ranu540
We want to use as many $99$s as possible. Then $99 \cdot 20 + 44 = 2024$, so our answer is $20 + 1 = \boxed{\textbf{(B) }21}$.
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bobjoebilly
79 posts
#4 • 1 Y
Y by pog
fairly certain this was also #3 on 2024 AMC 12A
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Jonysun
34 posts
#6
Y by
Yeah it's $\fbox{B} \Rightarrow 21$.
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HonestCat
972 posts
#7
Y by
MAA looked at this and said “yeah, this is harder than p2”
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Sedro
5848 posts
#8
Y by
B confirmed.
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The_Reaperr
43 posts
#9
Y by
Yes it is B
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ElaineGu
388 posts
#10
Y by
Yup B. 99*21>2024
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MathRook7817
810 posts
#11
Y by
ceil(2024/99)
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Marcus_Zhang
980 posts
#12
Y by
legit read this one wrong as "2024 is the sum of two not necesarily distinct two-digit numbers"

rest in peace my reading skills
This post has been edited 1 time. Last edited by Marcus_Zhang, Nov 7, 2024, 5:13 PM
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LostDreams
144 posts
#13
Y by
Notice how it says "not necessarily distinct"
So yea just max out with bunch of 99s and you get B
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golden_star_123
208 posts
#15
Y by
This one was extremely easy and should have been #1.

This is $\lceil \frac{2024}{99} \rceil$, which is $\boxed{21}$.
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aryanpadarthi
4995 posts
#16
Y by
this should have been #2 or #1

ez
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SomeonecoolLovesMaths
3255 posts
#17
Y by
oh no I assumed them to be distinct :(
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wangzrpi
159 posts
#18
Y by
HonestCat wrote:
MAA looked at this and said “yeah, this is harder than p2”

didn't even read p2
B confirmed
this should be an amc8 problem
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OlympusHero
17020 posts
#19
Y by
May have misbubbled C for this question :(
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xHypotenuse
780 posts
#20
Y by
OlympusHero wrote:
May have misbubbled C for this question :(

Don't worry I think answer choices and questions were mixed up as an anti-cheat
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MC_ADe
183 posts
#21
Y by
sillied, said A cuz 99x20, forgot to +1 cuz of 44
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gicyuraok2
1059 posts
#22
Y by
quite a simple problem, just $99*20+44=2024$, but you can easily mess this one up
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andrewcheng
525 posts
#23
Y by
pog wrote:
The number $2024$ is written as the sum of not necessarily distinct two-digit numbers. What is the least number of two-digit numbers needed to write this sum?

$\textbf{(A) }20\qquad\textbf{(B) }21\qquad\textbf{(C) }22\qquad\textbf{(D) }23\qquad\textbf{(E) }24$

easier than P2 should be moved
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Countmath1
180 posts
#24
Y by
I pigeonholed this one. Answer is $\boxed{\textbf{(B)\ 21}}.$
This post has been edited 3 times. Last edited by Countmath1, Nov 7, 2024, 11:29 PM
Reason: again latex
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navier3072
119 posts
#25
Y by
This is really harder than 2 huh :wacko:
$\left \lceil \frac{2024}{99} \right \rceil = 21$
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lvlup
3 posts
#26
Y by
Marcus_Zhang wrote:
legit read this one wrong as "2024 is the sum of two not necesarily distinct two-digit numbers"

rest in peace my reading skills

yah same i had 2 reread it some times on the test
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iwastedmyusername
144 posts
#28
Y by
what is bro doing
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Apple_maths60
26 posts
#29
Y by
2024 =20×99 +44
So, total 21 two-digit number required
Ans:21(B)
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