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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.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. 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 events:
[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
[*]April 8th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS State Discussion
April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
Website to learn math
hawa   68
N 21 minutes ago by GallopingUnicorn45
Hi, I'm kinda curious what website do yall use to learn math, like i dont find any website thats fun to learn math
68 replies
hawa
Apr 9, 2025
GallopingUnicorn45
21 minutes ago
Can someone explain this one
hawa   10
N 6 hours ago by VivaanKam
Suppose n is the largest integer obtained by solving the following inequality:

3+9+18+30+...+n
n < 2021.
10 replies
hawa
Yesterday at 1:36 AM
VivaanKam
6 hours ago
Alcumus specs
YeohZY   5
N Yesterday at 7:23 PM by PikaPika999
Hi, can I ask about how Alcumus gives you points? (I mean the number on the red/orange/green/blue line on top that gains once you get correct answers. I'll just call it the score). I want to get my score up to 100, but I always can't and get stuck at like 98 or sth. Why is this so, and also how does alcumus make that "score" higher?
5 replies
YeohZY
Apr 27, 2025
PikaPika999
Yesterday at 7:23 PM
Math Problem I cant figure out how to do without bashing
equalsmc2   3
N Yesterday at 3:55 PM by NovaFrost
Hi,
I cant figure out how to do these 2 problems without bashing. Do you guys have any ideas for an elegant solution? Thank you!
Prob 1.
An RSM sports field has a square shape. Poles with letters M, A, T, H are located at the corners of the square (see the diagram). During warm up, a student starts at any pole, runs to another pole along a side of the square or across the field along diagonal MT (only in the direction from M to T), then runs to another pole along a side of the square or along diagonal MT, and so on. The student cannot repeat a run along the same side/diagonal of the square in the same direction. For instance, she cannot run from M to A twice, but she can run from M to A and at some point from A to M. How many different ways are there to complete the warm up that includes all nine possible runs (see the diagram)? One possible way is M-A-T-H-M-H-T-A-M-T (picture attached)

Prob 2.
In the expression 5@5@5@5@5 you replace each of the four @ symbols with either +, or, or x, or . You can insert one or more pairs of parentheses to control the order of operations. Find the second least whole number that CANNOT be the value of the resulting expression. For example, each of the numbers 25=5+5+5+5+5 and 605+(5+5)×5+5 can be the value of the resulting expression.

Prob 3. (This isnt bashing I don't understand how to do it though)
Suppose BC = 3AB in rectangle ABCD. Points E and F are on side BC such that BE = EF = FC. Compute the sum of the degree measures of the four angles EAB, EAF, EAC, EAD.

P.S. These are from an RSM olympiad. The answers are
3 replies
equalsmc2
Apr 6, 2025
NovaFrost
Yesterday at 3:55 PM
No more topics!
Hello friends
bibidi_skibidi   9
N Apr 16, 2025 by giratina3
Now unfortunately I don't know the difficulty of the problems posted here but I'll try to replicate:

Bob has 20 apples and 19 oranges. How many ways can he split the fruits between 7 people if each person must have at least 1 apple and 2 oranges?

After looking at the other posters I realized just how bashy this is

Also I can only edit this message for now since new AoPS users can only send 6 messages every day
9 replies
bibidi_skibidi
Apr 15, 2025
giratina3
Apr 16, 2025
Hello friends
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G H BBookmark kLocked kLocked NReply
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bibidi_skibidi
5 posts
#1 • 2 Y
Y by aidan0626, GodGodGodGodGoose
Now unfortunately I don't know the difficulty of the problems posted here but I'll try to replicate:

Bob has 20 apples and 19 oranges. How many ways can he split the fruits between 7 people if each person must have at least 1 apple and 2 oranges?

After looking at the other posters I realized just how bashy this is

Also I can only edit this message for now since new AoPS users can only send 6 messages every day
This post has been edited 1 time. Last edited by bibidi_skibidi, Apr 15, 2025, 4:19 AM
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aidan0626
1875 posts
#2
Y by
Hi :)
This is standard stars and bars, there are $\binom{19}{6}$ ways to split the apples and $\binom{11}{6}$ ways to split the oranges, and you just multiply those two together to get the final answer.
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Soupboy0
344 posts
#3
Y by
bro i got sniped :rotfl:
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maxamc
568 posts
#4
Y by
aidan0626 wrote:
Hi :)
This is standard stars and bars, there are $\binom{19}{6}$ ways to split the apples and $\binom{11}{6}$ ways to split the oranges, and you just multiply those two together to get the final answer.

Wolfram Alpha gives 12534984 as our final answer (impossible to compute)
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fluffyyyyyyy
1 post
#5
Y by
hello sabko
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Leeoz
178 posts
#6
Y by
maxamc wrote:
aidan0626 wrote:
Hi :)
This is standard stars and bars, there are $\binom{19}{6}$ ways to split the apples and $\binom{11}{6}$ ways to split the oranges, and you just multiply those two together to get the final answer.

Wolfram Alpha gives 12534984 as our final answer (impossible to compute)

its not "impossible" :P
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Thayaden
1361 posts
#7
Y by
bibidi_skibidi wrote:
Now unfortunately I don't know the difficulty of the problems posted here but I'll try to replicate:

Bob has 20 apples and 19 oranges. How many ways can he split the fruits between 7 people if each person must have at least 1 apple and 2 oranges?

After looking at the other posters I realized just how bashy this is

Also I can only edit this message for now since new AoPS users can only send 6 messages every day

Start by giving away the required fruit this leaves 18 fruit for grabs namely 13 ap0pels and 5 oranges we set up stars and bars for extras and get (13+6)! but yet no oranges is diffrent from another orange and there are 5! ways to arage oranges and 13! ways to arange appels and 6! ways to arange bars so I think our answer is $\boxed{\frac{(18+6)!}{5!13!6!}}$ is what I got
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maxamc
568 posts
#8
Y by
Thayaden wrote:
bibidi_skibidi wrote:
Now unfortunately I don't know the difficulty of the problems posted here but I'll try to replicate:

Bob has 20 apples and 19 oranges. How many ways can he split the fruits between 7 people if each person must have at least 1 apple and 2 oranges?

After looking at the other posters I realized just how bashy this is

Also I can only edit this message for now since new AoPS users can only send 6 messages every day

Start by giving away the required fruit this leaves 18 fruit for grabs namely 13 ap0pels and 5 oranges we set up stars and bars for extras and get (13+6)! but yet no oranges is diffrent from another orange and there are 5! ways to arage oranges and 13! ways to arange appels and 6! ways to arange bars so I think our answer is $\boxed{\frac{(18+6)!}{5!13!6!}}$ is what I got

Your answer is 1153218528, not equal to my Wolfram Alpha answer described above, unfortunately wrong I think.
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BS2012
1025 posts
#9
Y by
its just Solution sketch which is the same as the 2nd post
This post has been edited 3 times. Last edited by BS2012, Apr 16, 2025, 12:52 AM
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giratina3
495 posts
#10
Y by
This is not bashy at all if you know Stars and Bars. I would recommend reading the later sections of Intro to Counting and Probability for more information.
This post has been edited 1 time. Last edited by giratina3, Apr 16, 2025, 2:57 AM
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