ka May Highlights and 2025 AoPS Online Class Information
jlacosta0
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]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.
Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28
Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19
Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30
Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14
Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19
Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)
Intermediate: Grades 8-12
Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22
MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21
AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22
Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22
Given the number of suggestions we’ve been receiving, we’re transitioning to a suggestion form. If you have a suggestion for the AoPS website, please submit the Google Form: Suggestion Form
To keep all new suggestions together, any new suggestion threads posted will be deleted.
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Source: German TST 2004, IMO ShortList 2003, geometry problem 2
Three distinct points ,, and are fixed on a line in this order. Let be a circle passing through and whose center does not lie on the line . Denote by the intersection of the tangents to at and . Suppose meets the segment at . Prove that the intersection of the bisector of and the line does not depend on the choice of .
Let be a triangle with circumcentre . The points and are interior points of the sides and respectively. Let and be the midpoints of the segments and . respectively, and let be the circle passing through and . Suppose that the line is tangent to the circle . Prove that
To all AoPS users and admin,
Sometimes I came across the marathon(i .e number theory marathon, functional equation marathon etc) which allow access only after submitting log in request.There is no other way to access the question related to that marathons.It would be glad to open all private marathons publicly.Thank you!
hi just so u no im not trying 2 postfarm i just wanted 2 c how many aops admins i could get 2 friend me. if u dont like this then feel free 2 remove it
aops admins so far:
asuth_asuth
gmass
rrusczyk
sbarrack
thats all right now!
plz friend
again, if u dont like this then plz remove if u dont want it here
Is Halp! a bot? This user has been posting questions in nearly all of my AoPS classes when the user isn't a part of the class, and this user has 150k posts.
I got curious and decided to see if I can ban myself from my own blog.
can site admins give it back? it says site admins are the administrators of this blog
I honestly don't know where I come up with stuff like this
hi srry if this is in the rong place i didnt no where 2 put it i was wondering how u find a user? i tried using the search but they dont have any posts? dont no wat 2 do...
k Introducing myself at AoPS, and what's your magic wand?
asuth_asuth1193
NMay 23, 2025
by Penguin117
Hi!
I'm Andrew Sutherland. I'm the new Chief Product Officer at AoPS. As you may have read, Richard is retiring and Ben Kornell and I are working together to lead the company now. I'm leading all the software and digital stuff at AoPS. I just wanted to say hello and introduce myself! I'm really excited to be part of the special community that is AoPS.
Previously, I founded Quizlet as a 15-year-old high school student. I did Course 6 at MIT and then left to lead Quizlet full-time for a total of 14 years. I took a few years off and now I'm doing AoPS! I wrote more about all that on my blog: https://asuth.com/im-joining-aops
I have a question for all of you. If you could wave a magic wand, and change anything about AoPS, what would it be? All suggestions welcome! Thank you.
Hi! Because you can't reset your alcumus when your in classes can you email sheriff and let them reset it? My class already ended but I think that you have to wait until it completely disappears but I'm signing up for another class so I can't reset it. So can you email sheriff?
Excuse me!
for I am a frashman in aops.I wonder how I could get back to mark all my read.All topics in the forum became as if they had been marked.Would someone please help me deal with it???
we know that if where , then it has an even tens digit because the remainder will not effect the tens digit if it is less than 10, and if it is in mod 20, then any number times 20 has an even tens digit
note that 3 and 20 are relatively prime, so using euler's extention of FLT, (note that )
so
so we only need to go through m=0,1,2,3,4,5,6, and 7, and show that the remainders are all less than 10, then all the other powers are also proven
thus every remainder for a power of is 1,3,7, or 9
so the tens digit must be even because any number mod 20 where the remainder is less than 10 will have an even tens digit
It can be written in the form:-
It can easily be shown that if it is 1, then it is 1 mod 4, if it is 3 then it is 3 mod 4, if it is 7 it is 3 mod 4 and if it is 9 then it is 1 mod 4.
(We know that
so,
,so on and similarly the others.
Observe that 3^3=27...
Now after this unit's place of 3^n can be 1,3,9,7,....the cycle repeats, observe each leaves a rem of 0 or 2, which when added to 2 makes it even, thus proved.
We are going to look at 3^k mod 20. It repeats 3,9,7,1. Hence result.
Could you please elaborate.
I think Singular is referring to the fact that when each of those numbers is multiplied by 3, their tens digit is a 2. Hence the sum of the tens digits must be even is what he is arguing we can infer from that....not entirely sure if that's a rigorous argument.
we know that if where , then it has an even tens digit because the remainder will not effect the tens digit if it is less than 10, and if it is in mod 20, then any number times 20 has an even tens digit
note that 3 and 20 are relatively prime, so using euler's extention of FLT, (note that )
so
so we only need to go through m=0,1,2,3,4,5,6, and 7, and show that the remainders are all less than 10, then all the other powers are also proven
thus every remainder for a power of is 1,3,7, or 9
so the tens digit must be even because any number mod 20 where the remainder is less than 10 will have an even tens digit
Prove that the ten's digit of any power of 3 is even.
Solution: Suppose that the power of is . We'll do a quick induction on . First check that ,,, and
Suppose, for some , has an even ten's digit, say . So:
1) If the unit's digit of is 1, then ten's digit of
2) If the unit's digit of is 3, then ten's digit of
3) If the unit's digit of is 7, then ten's digit of
4) If the unit's digit of is 9, then ten's digit of
(Where, ). Hence, our Induction is complete and all powers of have an even ten's digit!
Really sorry to bump this trivial problem but I thought this approach was nice. there exists such that so hence ( is the floor function).Note that is the number formed by removing the last digit of and moreover this is even.The result follows
This post has been edited 3 times. Last edited by leibnitz, Mar 19, 2020, 12:11 PM
Whoops, I fell asleep while solving this. Here's the solution:
Notice that the last digit of every number in the form (where is a non-negative integer) goes in the pattern: .
The last digit will be when , the last digit will be when , the last digit will be when and the last digit will be when . We can separate this problem into 4 different cases.
Case 1: The last digit of is - This means that . can be written as where is a non-negative integer. So, . We want the ten's digit to be even, so: This is clearly true as . So we have proved our statement for the first case.
Case 2: The last digit of is - This means that . can be written as and we can proceed to do the same thing as Case 1 to get that which is true.
Case 3: The last digit of is - This means that . can be written as and we can proceed to do the same thing as Case 1 to get that which is also true.
Case 4: The last digit of is - This means that . can be written as and we can proceed to do the same thing as Case 1 to get that which is true as and .
We try induction.
Base case: ,
Trivial.
Now let's assume it works for . Proof for .
The ten's digit of is even, so multiplying it with will also result to an even number. So the rest of the role is played by the units digit. Observe the only possible units digits are . Multiplying them respectively yields . They too add an even number to the tenth position.As two even numbers add up to an even number, the resultant ten's digit is again even. Hence proved.
This post has been edited 1 time. Last edited by SomeonecoolLovesMaths, Sep 16, 2024, 6:25 PM