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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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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
Past PUMaC results
mathkiddus   1
N 4 minutes ago by lpieleanu
Does anybody know where we can find old PUMaC results/placing cutoffs
1 reply
mathkiddus
an hour ago
lpieleanu
4 minutes ago
study for AIME to qual for USJMO
Logicus14   0
15 minutes ago
hi

so this is my first year doing amc10 (9th grade) and I got a decent enough score (121.5) on amc10b and after some research I found out that I need to get AT LEAST 10 problems right to qualify for USJMO

did anyone get 10+ problems right? If you did how did you study? any general tips?


ty
0 replies
Logicus14
15 minutes ago
0 replies
Problem 2
evt917   47
N 38 minutes ago by MC_ADe
Source: 2024 AMC 12B #2 / AMC 10B #2
What is $10! - 7! \cdot 6!$?

$
\textbf{(A) }-120 \qquad
\textbf{(B) }0 \qquad
\textbf{(C) }120 \qquad
\textbf{(D) }600 \qquad
\textbf{(E) }720 \qquad
$
47 replies
evt917
Nov 13, 2024
MC_ADe
38 minutes ago
AMC Music Video - Orz to the Legends (AMC 10/12 Version)
megahertz13   21
N 38 minutes ago by vsarg
Orz to the Legends (Music, AMC version - MegaMath Channel original)

Get pumped up for the AMCs!

[youtube]https://www.youtube.com/watch?v=KmiNI00uo-s&ab_channel=MegaMathChannel[/youtube]

Join our discord! https://discord.gg/hh7vntTb2E

Math Problem (AMC 10/12): Let $M$ be the smallest positive integer satisfying the property that $M^6$ is a multiple of both $2024^2$ and $2025^3$. How many positive divisors does $M$ have?

James has a geometric series with first term $1$ and common ratio $0<r<1$. Andy multiplies the first term by $\frac{7}{5}$, and squares the common ratio. If the sum of the first series equals the sum of the second series, the common sum can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
21 replies
2 viewing
megahertz13
Oct 13, 2024
vsarg
38 minutes ago
No more topics!
sum of 7th powers
gracemoon124   20
N Today at 12:20 AM by lpieleanu
Source: 2024 AMC 10B #7
What is the remainder when $7^{2024}+7^{2025}+7^{2026}$ is divided by $19$?

$
\textbf{(A) }0 \qquad
\textbf{(B) }1 \qquad
\textbf{(C) }7 \qquad
\textbf{(D) }11 \qquad
\textbf{(E) }18 \qquad
$
20 replies
gracemoon124
Nov 13, 2024
lpieleanu
Today at 12:20 AM
sum of 7th powers
G H J
Source: 2024 AMC 10B #7
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gracemoon124
871 posts
#1
Y by
What is the remainder when $7^{2024}+7^{2025}+7^{2026}$ is divided by $19$?

$
\textbf{(A) }0 \qquad
\textbf{(B) }1 \qquad
\textbf{(C) }7 \qquad
\textbf{(D) }11 \qquad
\textbf{(E) }18 \qquad
$
This post has been edited 1 time. Last edited by LauraZed, Nov 13, 2024, 5:28 PM
Reason: changing to official wording, adding answer options, and filling out source field
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gladIasked
592 posts
#2
Y by
answer is $0$; factor out a $7^{2024}$, then $7^2+7+1\equiv 0 \pmod {19}$

i did what @below did in contest
This post has been edited 4 times. Last edited by gladIasked, Nov 13, 2024, 5:24 PM
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gracemoon124
871 posts
#3
Y by
what i did was $7^3\equiv 1\pmod{19}$ lol? so it's $7^2+7+1\equiv 0\pmod{19}$
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mathboy282
2966 posts
#4 • 1 Y
Y by beastEJ
cycled 7,11,1 mod 19, and because 2024 == 2 mod 3, we have mod 19 to be 11+1+7=19 mod 19 ==0 (A)
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MathRook7817
234 posts
#5
Y by
its just 0, find 7^4 mod 19, and work from there
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pingpongmerrily
2424 posts
#6 • 1 Y
Y by Sedro
MathRook7817 wrote:
its just 0, find 7^4 mod 19, and work from there

not necessary to bash that, just factor out $7^2+7+1=57$ which is divisible by $19$ ans is 0=A
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anticodon
9 posts
#7 • 1 Y
Y by wangzrpi
mathboy282 wrote:
cycled 7,11,1 mod 19, and because 2024 == 2 mod 3, we have mod 19 to be 11+1+7=19 mod 19 ==0 (A)

Note that 2024, 2025, and 2026 are 3 consecutive integers and the mod 19 residues of powers of 7 cycle back every 3 numbers. So you can just add the possibilities (residue of 1+11+7 mod 19) so answer is 0
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mathprodigy2011
58 posts
#8
Y by
omg I thought that said 10b #17. and if it was 17 I would've flipped cuz I took 12b
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happypi31415
660 posts
#9
Y by
I misread the 19 as a 17 and got clinical depression during the test
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anticodon
9 posts
#10
Y by
This should have been named "Lucky Sevens 2024"
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GreenBanana666
519 posts
#11
Y by
Kinda ez for a P.7. $7^{2024}+7^{2025}+7^{2026}=7^{2024}(1+7+7^2)=7^{2024}(1+7+7^2)=7^{2024}(57)=7^{2024}(3)(19).$ So $7^{2024}+7^{2025}+7^{2026}$ is divisible by $19$. Therefore the answer is $\boxed{A}$.
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andrewcheng
482 posts
#12
Y by
doing some quick modulo math we find 7^1=7 7^2=11 and 7^3=1 mod 19
notice that since 2024 2025 and 2026 are consecutive and 7^x mod 19 loops every 3 numbers we can simply add 7+1+11=19=0 mod 19
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Sabburi
201 posts
#13
Y by
The powers of seven look between 7,1,11 in some order, so the sum is 19, which is 0(mod 19), or A
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emailhy
149 posts
#14
Y by
gracemoon124 wrote:
What is the remainder when $7^{2024}+7^{2025}+7^{2026}$ is divided by $19$?

$ \textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }7 \qquad \textbf{(D) }11 \qquad \textbf{(E) }18 \qquad this was in amc 10 b $
also i just did it
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amaops1123
1644 posts
#15
Y by
I'm like the only person in my school who didn't silly this somehow
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pingpongmerrily
2424 posts
#16
Y by
how do you silly this do kids not learn to factor nowadays? what are we coming to? the world is ending!!!
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GreenBanana666
519 posts
#17
Y by
fr :skull: silling easy probs with no traps and annoying bashing is just a skill issue. Also, why does everyone use mod. They arent gonna test us on mod on a prob 7.
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amaops1123
1644 posts
#18
Y by
pingpongmerrily wrote:
how do you silly this do kids not learn to factor nowadays? what are we coming to? the world is ending!!!

Idk man
There's this one guy who got like 132 or smth around that range and this was one of his sillies
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aleyang
178 posts
#19
Y by
I used both finding a pattern and factoring to double check my answer.
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AbhayAttarde01
1133 posts
#20
Y by
factor it out
$7^{2024}(1+7+49)$
$1+7+49=57$
$57$ IS divisible by $19$
so our answer should be $\boxed{A:0}$
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lpieleanu
2248 posts
#21
Y by
Solution
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N Quick Reply
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