Happy Thanksgiving! Please note that there are no classes November 25th-December 1st.

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k a AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
There is still time to train for the November 6th and November 12th AMC 10A/12A and AMC 10B/12B, respectively! Enroll in our weekend seminars to be held on November 2nd and 3rd (listed below) and you will learn problem strategies, test taking techniques, and be able to take a full practice test! Note that the “B” seminars will have different material from the “A” seminars which were held in October.

[list][*]Special AMC 10 Problem Seminar B
[*]Special AMC 12 Problem Seminar B[/list]
For those who want to take a free practice test before the AMC 10/12 competitions, you can simulate a real competition experience by following this link. As you assess your performance on these exams, be sure to gather data!

[list][*]Which problems did you get right?
[list][*]Was the topic a strength (e.g. number theory, geometry, counting/probability, algebra)?
[*]How did you prepare?
[*]What was your confidence level?[/list]
[*]Which problems did you get wrong?
[list][list][*]Did you make an arithmetic error?
[*]Did you misread the problem?
[*]Did you have the foundational knowledge for the problem?
[*]Which topics require more fluency through practice (e.g. number theory, geometry, counting/probability, algebra)?
[*]Did you run out of time?[/list][/list]
Once you have analyzed the results with the above questions, you will have a plan of attack for future contests! BEST OF LUCK to all competitors at this year’s AMC 10 and AMC 12!

Did you know that the day after both the AMC 10A/12A and AMC 10B/12B you can join a free math jam where our AoPS team will go over the most interesting problems? Find the schedule below under “Mark your calendars”.

Mark your calendars for these upcoming free math jams!
[list][*]November 20th: Amherst College Info Session, 7:30 pm ET: Matt McGann, Dean of Admission and Financial Aid at Amherst College, and Nathan Pflueger, math professor at Amherst College, will host an info session exploring both Amherst College specifically and liberal arts colleges generally. Topics include opportunities in math, the admission process, and financial aid for both US and international students.
[*]November 7th: 2024 AMC 10/12 A Discussion, Thursday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 A, administered November 6. We will discuss some of the most interesting problems from each test!
[*]November 13th: 2024 AMC 10/12 B Discussion, Wednesday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 B, administered November 12. We will discuss some of the most interesting problems from each test![/list]
AoPS Spring classes are open for enrollment. Get a jump on the New Year 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!

Whether you've taken our classes at AoPS Online or AoPS Academies or reached the top echelons of our competition training with our Worldwide Online Olympiad Training (WOOT) program, you can now add your AoPS experience to the education section 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.

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0 replies
jlacosta
Nov 1, 2024
0 replies
reverse of legendre formula
We2592   5
N an hour ago by NamelyOrange
1) How to solve it or reverse the process of legendre formula
for what values of $n$ does $n!$ ends in $37$ zeros?

please help

I know that $V_{p}(n!)=\sum_{k=1}^{\infty}[\frac{n}{p^{k}}]$
then how to find $n!$ please
5 replies
We2592
Sunday at 11:06 AM
NamelyOrange
an hour ago
Buffed AIME FE (Bashy)
NamelyOrange   13
N an hour ago by NamelyOrange
How many $7$-tuples $(a_1,a_2,\cdots,a_7)$ are there such that $a_{a_{a_i}}=a_i$ for all integer $1\le i \le 7$?

(Inspired by 2024 OTIS mock AIME P4.

Yes, the answer is above 1000.

No, I did not copy the question. I changed the $a_{a_i}$ in the original (2 $a$s) to $a_{a_{a_i}}$ (3 $a$s), which makes the question a lot more complicated.)
13 replies
NamelyOrange
Nov 23, 2024
NamelyOrange
an hour ago
ratio of factorial sums
smartvong   2
N an hour ago by lbh_qys
For any positive integer $k$, let $k! = 1 \times 2 \times \cdot \times k$ be the product of the first $k$ positive integers.
Find the value of $$N = \frac{1+\sum^{344}_{k=1}(k\times k!)}{1+\sum^{343}_{k=1}(k\times k!)}.$$
2 replies
smartvong
Nov 19, 2024
lbh_qys
an hour ago
Strange Inequality
CluelessSquare   6
N an hour ago by sqing
Minimize $\frac{15}{x} - \frac{3}{4y}$ if $x^2 + y^2 + \frac{1}{x} + \frac{1}{y} = \frac{27}{4}$, and $x,y > 0$
6 replies
CluelessSquare
Sep 17, 2020
sqing
an hour ago
No more topics!
Number theory(kinda)
MathChallenger101   9
N Nov 20, 2024 by MathChallenger101
Determine all triples (a, b, c) of integers with the property

a+b+c=a^3+b^3+c^3=3
9 replies
MathChallenger101
Nov 10, 2024
MathChallenger101
Nov 20, 2024
Number theory(kinda)
G H J
G H BBookmark kLocked kLocked NReply
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MathChallenger101
38 posts
#1
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Determine all triples (a, b, c) of integers with the property

a+b+c=a^3+b^3+c^3=3
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EaZ_Shadow
398 posts
#2
Y by
I claim $(1,1,1)$ is the only solution because gap as $a,b,c$ approaches infinity is huge, meaning that $a^3>2(a-1)^3$. Therefore, we are done. Please help me make this more rigorous.
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MathChallenger101
38 posts
#3
Y by
EaZ_Shadow wrote:
I claim $(1,1,1)$ is the only solution because gap as $a,b,c$ approaches infinity is huge, meaning that $a^3>2(a-1)^3$. Therefore, we are done. Please help me make this more rigorous.

Thanks for the quick answer, I get what you are saying, ill try to look into it as soon as possible, right now it's 12.30 pm where im at so it'll take a bit
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aether_1729
33 posts
#4 • 1 Y
Y by MathChallenger101
https://en.wikipedia.org/wiki/Sums_of_three_cubes

This might be relevant? I might be mistaken but apparently finding $a, b, c \in \mathbb{Z}$ such that $a^3 + b^3 + c^3 = 3$ excluding small cases like $(a, b, c) = (1, 1, 1)$ and $(4, 4, -5)$ (and its permutations) is hard. In both of the trivial cases, we also have $a + b + c = 3$ so those work.
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Davdav1232
8 posts
#5
Y by
EaZ_Shadow wrote:
I claim $(1,1,1)$ is the only solution because gap as $a,b,c$ approaches infinity is huge, meaning that $a^3>2(a-1)^3$. Therefore, we are done. Please help me make this more rigorous.

This is completely incorrect. $a/(a-1)^3$ approaches 1 as $a$ goes to infinity.
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Davdav1232
8 posts
#6
Y by
$a=3-b-c$. By algebra, $(b+c)(b-3)(c-3)=8.$ we can check cases by noticing $b, c= -5, -1, 1, 2, 4, 5, 7, 11$ and $b+c=-8, -4, -2, -1, 1, 2, 4, 8$ and get that the only options are $b, c=(-5, 1), (-5, 4), (-5, 7), (-1, 5), (-1, 2), (1, -5), (1, 7), (2, -1), (4, -5), (4, 4), (5, -1), (7, -5), (7, 1)$. Only triplets that work are $(a, b, c)=(1, 1, 1), (4, 4, -5), (4, -5, 4), (-5, 4, 4)$
Z K Y
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MathChallenger101
38 posts
#7
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Davdav1232 wrote:
$a=3-b-c$. By algebra, $(b+c)(b-3)(c-3)=8.$ we can check cases by noticing $b, c= -5, -1, 1, 2, 4, 5, 7, 11$ and $b+c=-8, -4, -2, -1, 1, 2, 4, 8$ and get that the only options are $b, c=(-5, 1), (-5, 4), (-5, 7), (-1, 5), (-1, 2), (1, -5), (1, 7), (2, -1), (4, -5), (4, 4), (5, -1), (7, -5), (7, 1)$. Only triplets that work are $(a, b, c)=(1, 1, 1), (4, 4, -5), (4, -5, 4), (-5, 4, 4)$

I understand everything other than the second line, where you say (b+c)(b-3)(c-3)=8, mind if you demonstrate in detail. Thank you for your time!
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MathChallenger101
38 posts
#8
Y by
Nevermind, I did it
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EaZ_Shadow
398 posts
#9
Y by
EaZ_Shadow wrote:
I claim $(1,1,1)$ is the only solution because gap as $a,b,c$ approaches infinity is huge, meaning that $a^3>2(a-1)^3$. Therefore, we are done. Please help me make this more rigorous.

Oops sorry I got this wrong
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MathChallenger101
38 posts
#10
Y by
EaZ_Shadow wrote:
EaZ_Shadow wrote:
I claim $(1,1,1)$ is the only solution because gap as $a,b,c$ approaches infinity is huge, meaning that $a^3>2(a-1)^3$. Therefore, we are done. Please help me make this more rigorous.

Oops sorry I got this wrong

No worries, I forgot to point it out, nice ideea though
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