ka July Highlights and 2025 AoPS Online Class Information
jwelsh0
Jul 1, 2025
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
and train with the best! Please note that early bird pricing ends August 19th!
Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.
Our full course list for upcoming classes is below:
All classes start 7:30pm ET/4:30pm PT unless otherwise noted.
Prealgebra 2
Friday, Jul 25 - Nov 21
Sunday, Aug 17 - Dec 14
Tuesday, Sep 9 - Jan 13
Thursday, Sep 25 - Jan 29
Sunday, Oct 19 - Feb 22
Monday, Oct 27 - Mar 2
Wednesday, Nov 12 - Mar 18
Introduction to Algebra A
Tuesday, Jul 15 - Oct 28
Sunday, Aug 17 - Dec 14
Wednesday, Aug 27 - Dec 17
Friday, Sep 5 - Jan 16
Thursday, Sep 11 - Jan 15
Sunday, Sep 28 - Feb 1
Monday, Oct 6 - Feb 9
Tuesday, Oct 21 - Feb 24
Sunday, Nov 9 - Mar 15
Friday, Dec 5 - Apr 3
Introduction to Counting & Probability
Wednesday, Jul 2 - Sep 17
Sunday, Jul 27 - Oct 19
Monday, Aug 11 - Nov 3
Wednesday, Sep 3 - Nov 19
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Friday, Oct 3 - Jan 16
Sunday, Oct 19 - Jan 25
Tuesday, Nov 4 - Feb 10
Sunday, Dec 7 - Mar 8
Introduction to Number Theory
Tuesday, Jul 15 - Sep 30
Wednesday, Aug 13 - Oct 29
Friday, Sep 12 - Dec 12
Sunday, Oct 26 - Feb 1
Monday, Dec 1 - Mar 2
Introduction to Algebra B
Friday, Jul 18 - Nov 14
Thursday, Aug 7 - Nov 20
Monday, Aug 18 - Dec 15
Sunday, Sep 7 - Jan 11
Thursday, Sep 11 - Jan 15
Wednesday, Sep 24 - Jan 28
Sunday, Oct 26 - Mar 1
Tuesday, Nov 4 - Mar 10
Monday, Dec 1 - Mar 30
Introduction to Geometry
Monday, Jul 14 - Jan 19
Wednesday, Aug 13 - Feb 11
Tuesday, Aug 26 - Feb 24
Sunday, Sep 7 - Mar 8
Thursday, Sep 11 - Mar 12
Wednesday, Sep 24 - Mar 25
Sunday, Oct 26 - Apr 26
Monday, Nov 3 - May 4
Friday, Dec 5 - May 29
Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)
Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET)
Intermediate: Grades 8-12
Intermediate Algebra
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22
Friday, Aug 8 - Feb 20
Tuesday, Aug 26 - Feb 24
Sunday, Sep 28 - Mar 29
Wednesday, Oct 8 - Mar 8
Sunday, Nov 16 - May 17
Thursday, Dec 11 - Jun 4
Precalculus
Wednesday, Aug 6 - Jan 21
Tuesday, Sep 9 - Feb 24
Sunday, Sep 21 - Mar 8
Monday, Oct 20 - Apr 6
Sunday, Dec 14 - May 31
Advanced: Grades 9-12
Calculus
Sunday, Sep 7 - Mar 15
Wednesday, Sep 24 - Apr 1
Friday, Nov 14 - May 22
Contest Preparation: Grades 6-12
MATHCOUNTS/AMC 8 Basics
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 17 - Nov 9
Wednesday, Sep 3 - Nov 19
Tuesday, Sep 16 - Dec 9
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Oct 6 - Jan 12
Thursday, Oct 16 - Jan 22
Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!)
MATHCOUNTS/AMC 8 Advanced
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 17 - Nov 9
Tuesday, Aug 26 - Nov 11
Thursday, Sep 4 - Nov 20
Friday, Sep 12 - Dec 12
Monday, Sep 15 - Dec 8
Sunday, Oct 5 - Jan 11
Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!)
Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!)
AMC 10 Problem Series
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 10 - Nov 2
Thursday, Aug 14 - Oct 30
Tuesday, Aug 19 - Nov 4
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!)
Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)
AMC 10 Final Fives
Friday, Aug 15 - Sep 12
Sunday, Sep 7 - Sep 28
Tuesday, Sep 9 - Sep 30
Monday, Sep 22 - Oct 13
Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, Oct 8 - Oct 29
Thursday, Oct 9 - Oct 30
AMC 12 Problem Series
Wednesday, Aug 6 - Oct 22
Sunday, Aug 10 - Nov 2
Monday, Aug 18 - Nov 10
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)
AMC 12 Final Fives
Thursday, Sep 4 - Sep 25
Sunday, Sep 28 - Oct 19
Tuesday, Oct 7 - Oct 28
Let be a positive integer. A class of students run races, in each of which they are ranked with no draws. A student is eligible for a rating for positive integers and if they come in the top places in at least of the races. Their final score is the maximum possible value of across all ratings for which they are eligible.
Find the maximum possible sum of all the scores of the students.
Find the number of positive integer divisors of Answer
Solution
The product of five consecutive integers reminds us of Thus, we are motivated to divide by and then use hockey stick. We have that which, by the hockey stick identity, is equal to Thus, Hence, has positive integer divisors.
In chess, a bishop attacks any piece that is in the same diagonal. I am placing two bishops on a chessboard on different squares. Let be the number of possible placements there are in which they are attacking each other. Find the first three digits of .
Answer
656
Solution
Let a pair be a pair of bishops that are in the same diagonal. Without loss of generality, we will count the number of good pairs in which the bishop in the upper row is also to the left of the one in the lower row, then double it because of symmetry.
Consider the diagonal that contains the upper-left and lower-right squares of the grid. From this diagonal containing squares, we can choose any squares to form a good pair, which gives
Then, consider the two diagonals containing squares, one being above and the other being below the main diagonal. This gives us good pairs. Similarly, from the diagonals containing squares, we get Thus, multiplying by (because of the WLOG), By the hockey stick identity, Thus, the first digits of is
Source: 2025 Penchick Online Mathematical Olympiad Qualifying 1 II. #10
Let be a triangle. Consider the points as the feet of the altitudes from respectively, and its orthocenter which we suppose is the midpoint of . Let be the midpoint of , be the midpoint of , and
Prove that .
\textbf{Problem 14.} There are girls and boys. It is known that for any two boys and any two girls, there exists at least one boy--girl pair who do not know each other (mutually). Prove that the total number of boy--girl pairs who know each other (regardless of order) does not exceed
Let for any . Let and be the maximum and minimum values of , respectively. Find the product of and .
Answer
Solution
We begin by completing the square to rewrite the expression in terms of , which we can simplify as 1 later.
Evaluating , we know that the range of trigonometric values are from -1 to 1, but since it is squared, we only consider [0, 1] as it will only yield nonnegative values. Thus, our minimum and maximum values of are 0 and 1, respectively.
Obviously I suspect the only solution is . It's easy to prove that and , but other than that, and trying out modulo 121, I didn't get much far on this one ... I might be missing something trivial though :|
Obviously I suspect the only solution is . It's easy to prove that and , but other than that, and trying out modulo 121, I didn't get much far on this one ... I might be missing something trivial though
Y bynarutomath96, Adventure10, Mango247, and 3 other users
Make a substitution (as Valentin said it is easy to prove - look modulo ). So our equation will take form . Suppose that there is a prime which divides both and . Then but also - contradiction. Hence and both are -th powers. Because we must have . Now using binomial expansion we get (looking at ) and from it it is easy to see that we must have (because else - right side would be larger).
i remember seeing these type of diophantine equations being solved using algebraic number theory , and if congruences dont work then inequalities is my biggest bet. but thats just my opinion .
Of course you cannot solve this modulo 11, since there is a solution! You have to try at least modulo 121 or 32. I tried both these, and modulo 64 also. Nothing
I checked again and found those interesting stuffs
1. the endings of are always "","", "", "".(dunno how to prove this, maybe there's some exceptions)
2. the endings of are always "","", "", "".
(Appearently, x is a multiple of 4)
Instead of , it's more convenient to consider the so-called integral closure. For and squarefree, this is the same. But if , you consider . And if is not squarefree, you first divide through the largest square dividing .
If you don't want the integral closure, a bit more work is required (you consider the "conductor" [correct english word¿]).
It's conjectured that there are infinitely many such . A complete list is known for .
But instead of looking for a list of UFD's, you may look for a list of class numbers (class number UFD). See http://mathworld.wolfram.com/ClassNumber.html for example.
2^4*(2^(x-4)-1)=11*(11^(y-1)-1)
11^(y-1) always ends in 1. So, 11^(y-1)-1 is divisible by 10.
But LHS is not divisible by 10 except for x=4.
So, x=4, y=1 is the only possible solution.
2^4*(2^(x-4)-1)=11*(11^(y-1)-1)
11^(y-1) always ends in 1. So, 11^(y-1)-1 is divisible by 10.
But LHS is not divisible by 10 except for x=4.
So, x=4, y=1 is the only possible solution.
Actually both and are Euclidean rings, thus UFDs also, so Megus' solution above is correct
Just a small correction: is in fact not an UFD (e.g. ). But is, implying the same way of solution.
In there exist infinitely many of units (invertible elements of this UFD). So, we need to consider the case where is arbitrary unit. Megus' solution is incomplete. Ok?
Take modulo first to see that we need, for . Then we must have . Similarly modulo gives . Substituting this becomes, Then taking modulo we find, Then this rearranges to Also taking modulo we have which requires . Now it is easy to see that there is no why satisfying our modulo equation so we are done. For our only solutions is .
This post has been edited 1 time. Last edited by Shreyasharma, Dec 25, 2023, 7:04 PM
I claim that is the only solution. Clearly it works. Plugging in all also gives that no works.
Now, taking both sides mod , note that the RHS is always mod no matter what is, meaning that the LHS must also be mod . Using our knowledge of orders, we deduce that is mod if and only if . So let for some integer . We now have that Now for the sake of contradiction, assume that there's a solution with . Taking the LHS mod , since is a positive integer and , we have that . Therefore the LHS is mod , meaning that the RHS must also be mod .
Listing the powers of mod , we get which gives us that must be mod . Now we take the equation mod . Since is mod , we have that the LHS mod must be either or . However, listing the powers of mod , we get which gives us that whenever we have as mod , the RHS must be or mod , a contradiction to the or . Therefore there are no solutions where , meaning that is the only solution, finishing the problem.
This post has been edited 1 time. Last edited by peppapig_, Jan 4, 2024, 8:34 PM Reason: \dots
The only solution where is . Otherwise, we subtract 11 from both sides to get
Taking modulo 5 then modulo 64, we find and . However, substituting these values back in, we find the equation has no solutions modulo 17 when .
We claim that the only solution is , which clearly works. We can quickly check that having or fails, so for what follows, we will assume and and show that this is impossible.
We must have so . We also require that , so we must have . Putting these together we have that .
Now we rewrite the equation as . Since the left side is , we require . After computing a few powers of , we can determine that we must have . Since , we know that divides the left side. By noting that is a factor of , we know that must divide the left side and thus also the right side, so we need . Using the fact that to simplify our computations, we can determine that the order of is and thus . However, we also require , contradiction.
Taking mod gives so Letting gives Taking mod gives so Write If we get the solution . Now assume Taking mod gives After experimenting some values, we see Since has a order of mod (we can check but ), let Taking mod gives which simplifies to (note that )Either possibility gives a contradiction, so the only solution is
We claim that the only solution is . It can be manually checked that all solutions with won't work, so from now we will assume that . First, taking mod 5 gives us that so . Therefore, we can rewrite as . Now, taking gives us This means that by binomial theorem, so must be odd. Therefore, we can let .
Now, combining all of our results so far, we have . Taking mod 17 now gives us Therefore, we have . When is positive, there are no solutions, but when is negative, we have . Therefore, now, we can write .
However, note that the order of is , so we have that contradicting the earlier , so we are done.
This post has been edited 1 time. Last edited by happypi31415, Sep 15, 2024, 6:17 PM
We try all up to , finding that the only pair is . Next, we take mod . For , we find that the term drops, leaving We can solve this, getting that . Next, we take mod (\textit{For my reference: FLT gives }). This gives that by FLT. This means we only need to try , which doesn't give a valid solution (for ). Thus, our answer is .