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k a January Highlights and 2025 AoPS Online Class Information
jlacosta 0
Jan 1, 2025
Happy New Year!!! Did you know, 2025 is the first perfect square year that any AoPS student has experienced? The last perfect square year was 1936 and the next one will be 2116! Let’s make it a perfect year all around by tackling new challenges, connecting with more problem-solvers, and staying curious!
We have some fun new things happening at AoPS in 2025 with new courses, such as self-paced Introduction to Algebra B, more coding, more physics, and, well, more!
There are a number of upcoming events, so be sure to mark your calendars for the following:
[list][*]Accelerated AIME Problem Series classes start on January 6th and 7th. These AIME classes will run three times a week throughout the month of January. With this accelerated track, you can fit three months of contest tips and training into four weeks finishing right in time for the AIME I on February 6th.
[*]Join our Math Jam on January 7th to learn about our Spring course options. We'll work through a few sample problems, discuss how the courses work, and answer your questions.
[*]RSVP for our New Year, New Challenges webinar on January 9th. We’ll discuss how you can meet your goals, useful strategies for your problem solving journey, and what classes and resources are available.
Have questions? Our Academic Success team is only an email away, write to us at success@aops.com.[/list]
AoPS Spring classes are open for enrollment. Get a jump on 2025 and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!
Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile! Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!
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.
Introductory: Grades 5-10
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Introduction to Algebra A
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Introduction to Counting & Probability Self-Paced
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Tuesday, Jan 28 - Apr 15
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Sat & Sun, Jan 11 - Jan 12 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)
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Sunday, Feb 23 - May 11
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AIME Problem Series A
Tue, Thurs & Sun, Jan 7 - Feb (meets three times each week!)
AIME Problem Series B
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Special AIME Problem Seminar A
Sat & Sun, Jan 25 - Jan 26 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)
Special AIME Problem Seminar B
Sat & Sun, Feb 1 - Feb 2 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)
F=ma Problem Series
Wednesday, Feb 19 - May 7
Programming
Introduction to Programming with Python
Friday, Jan 17 - Apr 4
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Tuesday, Feb 25 - May 13
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Introduction to Physics
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Sat & Sun, Dec 14 - Dec 15 (4:00 - 7:00 pm ET/1:00 - 4:00pm PT)
We have some fun new things happening at AoPS in 2025 with new courses, such as self-paced Introduction to Algebra B, more coding, more physics, and, well, more!
There are a number of upcoming events, so be sure to mark your calendars for the following:
[list][*]Accelerated AIME Problem Series classes start on January 6th and 7th. These AIME classes will run three times a week throughout the month of January. With this accelerated track, you can fit three months of contest tips and training into four weeks finishing right in time for the AIME I on February 6th.
[*]Join our Math Jam on January 7th to learn about our Spring course options. We'll work through a few sample problems, discuss how the courses work, and answer your questions.
[*]RSVP for our New Year, New Challenges webinar on January 9th. We’ll discuss how you can meet your goals, useful strategies for your problem solving journey, and what classes and resources are available.
Have questions? Our Academic Success team is only an email away, write to us at success@aops.com.[/list]
AoPS Spring classes are open for enrollment. Get a jump on 2025 and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!
Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile! Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!
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.
Introductory: Grades 5-10
Prealgebra 1
Sunday, Jan 5 - Apr 20
Wednesday, Jan 15 - Apr 30
Monday, Feb 3 - May 19
Sunday, Mar 2 - Jun 22
Friday, Mar 28 - Jul 18
Sunday, Apr 13 - Aug 10
Prealgebra 1 Self-Paced
Prealgebra 2
Wednesday, Jan 8 - Apr 23
Sunday, Jan 19 - May 4 (1:00 - 2:15 pm ET/10:00 - 11:15 am PT)
Monday, Jan 27 - May 12
Tuesday, Jan 28 - May 13 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT)
Sunday, Feb 16 - Jun 8
Tuesday, Mar 25 - Jul 8
Sunday, Apr 13 - Aug 10
Prealgebra 2 Self-Paced
Introduction to Algebra A
Tuesday, Jan 7 - Apr 22
Wednesday, Jan 29 - May 14
Sunday, Feb 16 - Jun 8 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)
Sunday, Mar 23 - Jul 20
Monday, Apr 7 - Jul 28
Introduction to Algebra A Self-Paced
Introduction to Counting & Probability
Wednesday, Jan 8 - Mar 26
Thursday, Jan 30 - Apr 17
Sunday, Feb 9 - Apr 27 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)
Sunday, Mar 16 - Jun 8
Wednesday, Apr 16 - Jul 2
Introduction to Counting & Probability Self-Paced
Introduction to Number Theory
Tuesday, Jan 28 - Apr 15
Sunday, Feb 16 - May 4
Monday, Mar 17 - Jun 9
Thursday, Apr 17 - Jul 3
Introduction to Algebra B
Tuesday, Jan 28 - May 13
Thursday, Feb 13 - May 29
Sunday, Mar 2 - Jun 22
Monday, Mar 17 - Jul 7
Wednesday, Apr 16 - Jul 30
Introduction to Geometry
Wednesday, Jan 8 - Jun 18
Thursday, Jan 30 - Jul 10
Friday, Feb 14 - Aug 1
Tuesday, Mar 4 - Aug 12
Sunday, Mar 23 - Sep 21
Wednesday, Apr 23 - Oct 1
Intermediate: Grades 8-12
Intermediate Algebra
Friday, Jan 17 - Jun 27
Wednesday, Feb 12 - Jul 23
Sunday, Mar 16 - Sep 14
Tuesday, Mar 25 - Sep 2
Monday, Apr 21 - Oct 13
Intermediate Counting & Probability
Monday, Feb 10 - Jun 16
Sunday, Mar 23 - Aug 3
Intermediate Number Theory
Thursday, Feb 20 - May 8
Friday, Apr 11 - Jun 27
Precalculus
Wednesday, Jan 8 - Jun 4
Tuesday, Feb 25 - Jul 22
Sunday, Mar 16 - Aug 24
Wednesday, Apr 9 - Sep 3
Advanced: Grades 9-12
Olympiad Geometry
Wednesday, Mar 5 - May 21
Calculus
Friday, Feb 28 - Aug 22
Sunday, Mar 30 - Oct 5
Contest Preparation: Grades 6-12
MATHCOUNTS/AMC 8 Basics
Tuesday, Feb 4 - Apr 22
Sunday, Mar 23 - Jun 15
Wednesday, Apr 16 - Jul 2
MATHCOUNTS/AMC 8 Advanced
Sunday, Feb 16 - May 4
Friday, Apr 11 - Jun 27
Special AMC 8 Problem Seminar A
Sat & Sun, Jan 11 - Jan 12 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)
Special AMC 8 Problem Seminar B
Sat & Sun, Jan 18 - Jan 19 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)
AMC 10 Problem Series
Sunday, Feb 9 - Apr 27
Tuesday, Mar 4 - May 20
Monday, Mar 31 - Jun 23
AMC 10 Final Fives
Sunday, Feb 9 - Mar 2 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)
AMC 12 Problem Series
Sunday, Feb 23 - May 11
AMC 12 Final Fives
Sunday, Feb 9 - Mar 2 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)
AIME Problem Series A
Tue, Thurs & Sun, Jan 7 - Feb (meets three times each week!)
AIME Problem Series B
Mon, Wed & Fri, Jan 6 - Jan 31 (meets three times each week!)
Special AIME Problem Seminar A
Sat & Sun, Jan 25 - Jan 26 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)
Special AIME Problem Seminar B
Sat & Sun, Feb 1 - Feb 2 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)
F=ma Problem Series
Wednesday, Feb 19 - May 7
Programming
Introduction to Programming with Python
Friday, Jan 17 - Apr 4
Sunday, Feb 16 - May 4
Monday, Mar 24 - Jun 16
Intermediate Programming with Python
Tuesday, Feb 25 - May 13
USACO Bronze Problem Series
Sunday, Jan 5 - Mar 23
Thursday, Feb 6 - Apr 24
Physics
Introduction to Physics
Friday, Feb 7 - Apr 25
Sunday, Mar 30 - Jun 22
Physics 1: Mechanics
Sunday, Feb 9 - Aug 3
Tuesday, Mar 25 - Sep 2
Relativity
Sat & Sun, Dec 14 - Dec 15 (4:00 - 7:00 pm ET/1:00 - 4:00pm PT)
0 replies
Exercise 1.1
Alohomora. 0
Dec 10, 2024
What I've tried so far:
<Describe what you have tried so far here. That way, we can do a better job helping you!>
Let be the greatest number in the finite set . Take an arbitrary number in . We have: So is not prime and therefore doesn't have any prime.
Where I'm stuck:
<Describe what's confusing you, or what your question is here!>
<Describe what you have tried so far here. That way, we can do a better job helping you!>
Let be the greatest number in the finite set . Take an arbitrary number in . We have: So is not prime and therefore doesn't have any prime.
Where I'm stuck:
<Describe what's confusing you, or what your question is here!>
0 replies
Exercise 0.4
Thayaden 3
N
Nov 9, 2024
by felixgotti
In general, is a ring. A ring is and integral domain iff is the only zero-divisor. Another way of putting this is for some prime number show that,
where where . Consider for some where is still some prime number. This implies although by definition ONLY for some this implies that either or although in modulus we have thus or must be a zero-divisor thus we can conclude that for any given prime is an integral domain. Consider if we have that implies that , in this case, or where is equal to the other that implies that is a zero divisor and if is non-prime that implies that for some non- divisor, is a zero divisor.
where where . Consider for some where is still some prime number. This implies although by definition ONLY for some this implies that either or although in modulus we have thus or must be a zero-divisor thus we can conclude that for any given prime is an integral domain. Consider if we have that implies that , in this case, or where is equal to the other that implies that is a zero divisor and if is non-prime that implies that for some non- divisor, is a zero divisor.
3 replies
Exercise 0.3
Thayaden 2
N
Nov 7, 2024
by Thayaden
Crazy Idea but I think the condtions work
and
yet
and
yet
2 replies
Exercise 0.8
Thayaden 1
N
Oct 31, 2024
by felixgotti
Consider and such that we clearly see that,
Recall thus we can clearly see that as they are each other inverse that thus let,
Thus we see,
In other words and are elements in the group of units of although since it is also the group of units of therefor
Recall thus we can clearly see that as they are each other inverse that thus let,
Thus we see,
In other words and are elements in the group of units of although since it is also the group of units of therefor
1 reply
units...
Thayaden 1
N
Oct 31, 2024
by felixgotti
Part 1:
Notice thus for we might enharite this too. This might be easily proven by letting the coefficient of the unreal part be . Thus we might know that at least is a unit. Taking that idea once again letting the real part be we have,
Thus as we are communitive is in the set this makes sense from an algebraic perspective on the complex plane when multiplying a unit we have a change of direction likewise this is also true for real numbers thus
Notice thus for we might enharite this too. This might be easily proven by letting the coefficient of the unreal part be . Thus we might know that at least is a unit. Taking that idea once again letting the real part be we have,
Thus as we are communitive is in the set this makes sense from an algebraic perspective on the complex plane when multiplying a unit we have a change of direction likewise this is also true for real numbers thus
1 reply
Exercise 0.5 crazy
Thayaden 1
N
Oct 31, 2024
by felixgotti
Let be a finite integral domain that is not a field. For some there is no for some many as is not a field. Consider the powers of ,
let,
Consider the first powers as unique, implying that is not unique or consider the first as not unique. In any given case for some ,
If that implies that (this can be shown inductively) and we know that so that implies that,
Although that implies that has an inverse, that begins the only numbers that could not have an inverse indeed do have an inverse thus all numbers in have an inverse thus is a field!
let,
Consider the first powers as unique, implying that is not unique or consider the first as not unique. In any given case for some ,
If that implies that (this can be shown inductively) and we know that so that implies that,
Although that implies that has an inverse, that begins the only numbers that could not have an inverse indeed do have an inverse thus all numbers in have an inverse thus is a field!
1 reply
Exercise 0.7
Thayaden 0
Oct 2, 2024
If is an intergral domain then is too,
Let such that they are non zero, denote
Taking let be the leading coefficient of . Notice that thus so in short for the leading coefficient of cannot be thus must be an integral domain.
Going the other way say is an integral domain,
For the sake of contradiction let such that and , let taking there product,
This contradicts our previous statement thus is an integral domain!
Finally iff is an integral domain then must also be an integral domain
Let such that they are non zero, denote
Taking let be the leading coefficient of . Notice that thus so in short for the leading coefficient of cannot be thus must be an integral domain.
Going the other way say is an integral domain,
For the sake of contradiction let such that and , let taking there product,
This contradicts our previous statement thus is an integral domain!
Finally iff is an integral domain then must also be an integral domain
0 replies
Exercise 0.2
Thayaden 0
Sep 16, 2024
Let be a submonoid of , where runs over is the index set. Let be the intersection of all possible submonoids of . Notice now we just need to prove . Let be the identity of notice that and in turn therfor has the identity of the first characteristic of a submonoid. Now we just need to prove that it is a monoid with respect to the operation of . Let by intersection rules and and notice is a submonoid in itself and thus the operation for every thus by intersection rules again. Thus is a submonoid in itself!
Let be a subgroup of , where runs over is the index set. Let be the intersection of all possible subgroups of . Notice now we just need to prove . Let be the identity of notice that and in turn therfor has the identity of the first characteristic of a subgroup. Now we need to prove that it is a subgroup with respect to the operation of . Let by intersection rules and and notice is a subgroup in itself and thus the operation for every thus by intersection rules again. Finally, we need to show that every element has an inverse. Let and let represent the inverse of . Notice if then it follows that although notice that all are subgroups of and thus and since for any then by intersection thus is a subgroup of
Let be a subgroup of , where runs over is the index set. Let be the intersection of all possible subgroups of . Notice now we just need to prove . Let be the identity of notice that and in turn therfor has the identity of the first characteristic of a subgroup. Now we need to prove that it is a subgroup with respect to the operation of . Let by intersection rules and and notice is a subgroup in itself and thus the operation for every thus by intersection rules again. Finally, we need to show that every element has an inverse. Let and let represent the inverse of . Notice if then it follows that although notice that all are subgroups of and thus and since for any then by intersection thus is a subgroup of
0 replies
Exercise 0.1(part 1)
Thayaden 1
N
Sep 11, 2024
by Thayaden
let by definition,
this is true for all thus let this might lead us to (from ),
Now let (using ),
By the transitive property,
as the first condition is true, we might then say that given all identity elements of denoted as we might have,
and inductively,
thus,
. Or in plain text, we might say that as one identity is equal to another then if we have some identity it is the same as the identity we already have.
this is true for all thus let this might lead us to (from ),
Now let (using ),
By the transitive property,
as the first condition is true, we might then say that given all identity elements of denoted as we might have,
and inductively,
thus,
. Or in plain text, we might say that as one identity is equal to another then if we have some identity it is the same as the identity we already have.
1 reply
Exercise 3.4
FoxProdigy 1
N
Aug 19, 2024
by felixgotti
Problem Statement: Prove that a monoid is a GL-monoid if and only if it is prime-like.
Proof:
Prime-like GL:
Let be a prime-like monoid. Suppose such that . Suppose further towards a contradiction that ; that is, that there exists some divisor with . Then because is prime-like, has some divisor with either or . Either way, we also must have , which means that either or . This is a contradiction. Thus, , which means is GL.
GL prime-like:
Let be a GL monoid, and suppose with . Assume towards a contradiction that for all , and (in other words, shares no divisors with or ). This implies that . Since is GL, this means that . This contradicts the assumption that . Thus, there must exist some with or , which means is prime-like.
Proof:
Prime-like GL:
Let be a prime-like monoid. Suppose such that . Suppose further towards a contradiction that ; that is, that there exists some divisor with . Then because is prime-like, has some divisor with either or . Either way, we also must have , which means that either or . This is a contradiction. Thus, , which means is GL.
GL prime-like:
Let be a GL monoid, and suppose with . Assume towards a contradiction that for all , and (in other words, shares no divisors with or ). This implies that . Since is GL, this means that . This contradicts the assumption that . Thus, there must exist some with or , which means is prime-like.
1 reply