ka March Highlights and 2025 AoPS Online Class Information
jlacosta0
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]
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.
Prealgebra 1
Sunday, Mar 2 - Jun 22
Friday, Mar 28 - Jul 18
Sunday, Apr 13 - Aug 10
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29
Prealgebra 2
Tuesday, Mar 25 - Jul 8
Sunday, Apr 13 - Aug 10
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21
Introduction to Algebra A
Sunday, Mar 23 - Jul 20
Monday, Apr 7 - Jul 28
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
Sunday, Mar 16 - Jun 8
Wednesday, Apr 16 - Jul 2
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
Monday, Mar 17 - Jun 9
Thursday, Apr 17 - Jul 3
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
Sunday, Mar 2 - Jun 22
Wednesday, Apr 16 - Jul 30
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14
Introduction to Geometry
Tuesday, Mar 4 - Aug 12
Sunday, Mar 23 - Sep 21
Wednesday, Apr 23 - Oct 1
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
Intermediate: Grades 8-12
Intermediate Algebra
Sunday, Mar 16 - Sep 14
Tuesday, Mar 25 - Sep 2
Monday, Apr 21 - Oct 13
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
Sunday, Mar 23 - Jun 15
Wednesday, Apr 16 - Jul 2
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
Friday, Apr 11 - Jun 27
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
Tuesday, Mar 4 - May 20
Monday, Mar 31 - Jun 23
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
Monday, Mar 24 - Jun 16
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
Sketch:
Clearly is a rational number, moreover, is of the form . Now after considering cases of what residue gives modulo , we're about to get equations.
This problem seems unanswered yet, so i will write my solution. Corrections are always welcome. Thanks
Notice that the problem is equivalent to
Also, notice that
This gives
So,
Now, consider the function . We will find on which interval does increase, so we need to take derivative, then , then .
Notice again that . Since , and is increasing on , then , so .
Noticing that and using similar reasoning, gives .
So, we have
This means, the possible values for are
If , then , so and .
So, .
Plugging each value to , we see that the only possible values are .
If , then , so and .
Checking, we see that is the only possible value for in this case.
If , then , so and .
Thus, .
Checking again, we see that the possible values are .
So
Hence, since , then we substitute those values there and get :
Y bytongzhao, ZacPower123, vsathiam, Adventure10, Mango247, and 1 other user
Holddddd up if you're saying I spent 30 minutes on this freaking problem just to get it wrong at midnight I will quite possibly fall into depression and eat all of my halloween candy.
First off doesn't even work. Nor does and (trust me I calculated each of those individually to check for any mistakes).
Here is my solution:
The restriction of tells us that . It also tells us that must be an integer which means that is an integer.
This means that for some integer .
First let's investigate
By rearranging the numbers we have . By RRT we find that is a root of . Factoring we get .
By the quadratic equation the solutions for are .
Thus this condition is satisfied if or
Now let's investigate
We have that when but when . Obviously increases faster than so . Note that this bound is not strong and in fact for some
Now we want to find the values of in the range such that for some integer .
Note that . Thus
Thus the possible values of in this range are which we can check to see that they both work.
Now we approximate
Thus the possible values of this range are and . But wait! We have to check if these values indeed satisfy because we never determined the value such that . Plugging in we see that the condition is not satisfied. As again, grows faster than won't work either.
Thus the solutions are simply
Hm it's too late right now (even with daylight savings) for me to have the energy to read the above solution so I might do that later and try finding what's wrong.
I don't think there's anything wrong in MrRTI's solution. He just gave a correct proof for reducing the set of possible solutions to the 9 numbers he stated as "solutions". The mistake is just that not all these numbers really satisfy the equation because the conditions you elaborated are just necessary but not sufficient ones. So, you just need to plug in all your 9 values to see that only and will be solutions.
Holddddd up if you're saying I spent 30 minutes on this freaking problem just to get it wrong at midnight I will quite possibly fall into depression and eat all of my halloween candy.
First off doesn't even work. Nor does and (trust me I calculated each of those individually to check for any mistakes).
Here is my solution:
The restriction of tells us that . It also tells us that must be an integer which means that is an integer.
This means that for some integer .
First let's investigate
By rearranging the numbers we have . By RRT we find that is a root of . Factoring we get .
By the quadratic equation the solutions for are .
Thus this condition is satisfied if or
Now let's investigate
We have that when but when . Obviously increases faster than so . Note that this bound is not strong and in fact for some
Now we want to find the values of in the range such that for some integer .
Note that . Thus
Thus the possible values of in this range are which we can check to see that they both work.
Now we approximate
Thus the possible values of this range are and . But wait! We have to check if these values indeed satisfy because we never determined the value such that . Plugging in we see that the condition is not satisfied. As again, grows faster than won't work either.
Thus the solutions are simply
Hm it's too late right now (even with daylight savings) for me to have the energy to read the above solution so I might do that later and try finding what's wrong.
efang's right. Also, you could have noted that the fractional part of x lies in the set S = {} since 4x+3 must be an integer. From there you note that , where x = n+s, n is an integer, and s S. Separating the bound into two polynomials and casework bashing gives you x = -1.25 and -1.
I think I have a somehow easier to understand (for people more in the beginners level like me) solution that isn't too long why do I always induct inequalities like at JBMO
*Idk how to type floor function, so absolute value(||) here means floor function
Claim 1: For , we have
Proof by Induction: Clearly, for x element of (I shud really learn latex better...),
We have
Suppose it is true for x element of , where k is a positive integer that is at least 3
Then when x element of , we have (Since it is true for x=k) (As ) (Again, since )
Thus, the Claim is proved
Claim 2:For , we have
Proof: Let
It suffices to show by flipping the signs, where
But this follows immediately from Claim 1
Thus, the Claim is proved
Hence, by these 2 Claims, x must be an element of
Clearly, the LHS is an integer, so 4x+3, and hence, 4x must be an integer
So x must be of the form , where k is an integer
Combining this with the range, we get that there's only 23 possible values
Checking (a lot of them can be eliminated in groups by a bit of bounding),
we see that are the only solutions
.
Brb. Ok done
This post has been edited 4 times. Last edited by Devastator, Jun 2, 2018, 12:16 PM Reason: Yes