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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 May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/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.

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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
[TEST RELEASED] OMMC Year 5
DottedCaculator   140
N 4 minutes ago by BS2012
Test portal: https://ommc-test-portal-2025.vercel.app/

Hello to all creative problem solvers,

Do you want to work on a fun, untimed team math competition with amazing questions by MOPpers and IMO & EGMO medalists? $\phantom{You lost the game.}$
Do you want to have a chance to win thousands in cash and raffle prizes (no matter your skill level)?

Check out the fifth annual iteration of the

Online Monmouth Math Competition!

Online Monmouth Math Competition, or OMMC, is a 501c3 accredited nonprofit organization managed by adults, college students, and high schoolers which aims to give talented high school and middle school students an exciting way to develop their skills in mathematics.

Our website: https://www.ommcofficial.org/

This is not a local competition; any student 18 or younger anywhere in the world can attend. We have changed some elements of our contest format, so read carefully and thoroughly. Join our Discord or monitor this thread for updates and test releases.

How hard is it?

We plan to raffle out a TON of prizes over all competitors regardless of performance. So just submit: a few minutes of your time will give you a great chance to win amazing prizes!

How are the problems?

You can check out our past problems and sample problems here:
https://www.ommcofficial.org/sample
https://www.ommcofficial.org/2022-documents
https://www.ommcofficial.org/2023-documents
https://www.ommcofficial.org/ommc-amc

How will the test be held?/How do I sign up?

Solo teams?

Test Policy

Timeline:
Main Round: May 17th - May 24th
Test Portal Released. The Main Round of the contest is held. The Main Round consists of 25 questions that each have a numerical answer. Teams will have the entire time interval to work on the questions. They can submit any time during the interval. Teams are free to edit their submissions before the period ends, even after they submit.

Final Round: May 26th - May 28th
The top placing teams will qualify for this invitational round (5-10 questions). The final round consists of 5-10 proof questions. Teams again will have the entire time interval to work on these questions and can submit their proofs any time during this interval. Teams are free to edit their submissions before the period ends, even after they submit.

Conclusion of Competition: Early June
Solutions will be released, winners announced, and prizes sent out to winners.

Scoring:

Prizes:

I have more questions. Whom do I ask?

We hope for your participation, and good luck!

OMMC staff

OMMC’S 2025 EVENTS ARE SPONSORED BY:

[list]
[*]Nontrivial Fellowship
[*]Citadel
[*]SPARC
[*]Jane Street
[*]And counting!
[/list]
140 replies
+6 w
DottedCaculator
Apr 26, 2025
BS2012
4 minutes ago
Another right angled triangle
ariopro1387   0
an hour ago
Source: Iran Team selection test 2025 - P7
Let $ABC$ be a right angled triangle with $\angle A=90$. Point $M$ is the midpoint of side $BC$ And $P$ be an arbitrary point on $AM$. The reflection of $BP$ over $AB$ intersects lines $AC$ and $AM$ at $T$ and $Q$, respectively. The circumcircles of $BPQ$ and $ABC$ intersect again at $F$. Prove that the center of the circumcircle of $CFT$ lies on $BQ$.
0 replies
ariopro1387
an hour ago
0 replies
diophantine equation
m4thbl3nd3r   1
N an hour ago by whwlqkd
Find all positive integers $n,k$ such that $$5^{2n+1}-5^n+1=k^2$$
1 reply
m4thbl3nd3r
Today at 10:34 AM
whwlqkd
an hour ago
[CASH PRIZES] IndyINTEGIRLS Spring Math Competition
Indy_Integirls   58
N an hour ago by PeaceLilly22
[center]IMAGE

Greetings, AoPS! IndyINTEGIRLS will be hosting a virtual math competition on May 25,
2024 from 12 PM to 3 PM EST.
Join other woman-identifying and/or non-binary "STEMinists" in solving problems, socializing, playing games, winning prizes, and more! If you are interested in competing, please register here![/center]

----------

[center]Important Information[/center]

Eligibility: This competition is open to all woman-identifying and non-binary students in middle and high school. Non-Indiana residents and international students are welcome as well!

Format: There will be a middle school and high school division. In each separate division, there will be an individual round and a team round, where students are grouped into teams of 3-4 and collaboratively solve a set of difficult problems. There will also be a buzzer/countdown/Kahoot-style round, where students from both divisions are grouped together to compete in a MATHCOUNTS-style countdown round! There will be prizes for the top competitors in each division.

Problem Difficulty: Our amazing team of problem writers is working hard to ensure that there will be problems for problem-solvers of all levels! The middle school problems will range from MATHCOUNTS school round to AMC 10 level, while the high school problems will be for more advanced problem-solvers. The team round problems will cover various difficulty levels and are meant to be more difficult, while the countdown/buzzer/Kahoot round questions will be similar to MATHCOUNTS state to MATHCOUNTS Nationals countdown round in difficulty.

Platform: This contest will be held virtually through Zoom. All competitors are required to have their cameras turned on at all times unless they have a reason for otherwise. Proctors and volunteers will be monitoring students at all times to prevent cheating and to create a fair environment for all students.

Prizes: At this moment, prizes are TBD, and more information will be provided and attached to this post as the competition date approaches. Rest assured, IndyINTEGIRLS has historically given out very generous cash prizes, and we intend on maintaining this generosity into our Spring Competition.

Contact & Connect With Us: Email us at indy@integirls.org.

---------
[center]Help Us Out

Please help us in sharing the news of this competition! Our amazing team of officers has worked very hard to provide this educational opportunity to as many students as possible, and we would appreciate it if you could help us spread the word!
58 replies
Indy_Integirls
May 11, 2025
PeaceLilly22
an hour ago
ISI UGB 2025 P2
SomeonecoolLovesMaths   11
N an hour ago by ProMaskedVictor
Source: ISI UGB 2025 P2
If the interior angles of a triangle $ABC$ satisfy the equality, $$\sin ^2 A + \sin ^2 B + \sin^2  C = 2 \left( \cos ^2 A + \cos ^2 B + \cos ^2 C \right),$$prove that the triangle must have a right angle.
11 replies
SomeonecoolLovesMaths
May 11, 2025
ProMaskedVictor
an hour ago
Functional Inequaility
ariopro1387   2
N an hour ago by Triborg-V
Source: Own
Find all functions \(f: \mathbb{R} \rightarrow \mathbb{R}\) such that for any real numbers \(x\) and \(y\), the following inequality holds:
\[
f\left(x^2+2y f(x)\right) + (f(y))^2 \leq (f(x+y))^2
\]
2 replies
1 viewing
ariopro1387
Apr 9, 2025
Triborg-V
an hour ago
Problem 7
SlovEcience   4
N an hour ago by SlovEcience
Consider the sequence \((u_n)\) defined by \(u_0 = 5\) and
\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]a) Prove that there exist infinitely many positive integers \(n\) such that \(u_n > 2020n\).

b) Compute
\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]
4 replies
+1 w
SlovEcience
May 14, 2025
SlovEcience
an hour ago
Orthocentres of triangles ABC and AB’C’
Stun   40
N an hour ago by mathwiz_1207
Source: IMO Shortlist 1995, G8
Suppose that $ ABCD$ is a cyclic quadrilateral. Let $ E = AC\cap BD$ and $ F = AB\cap CD$. Denote by $ H_{1}$ and $ H_{2}$ the orthocenters of triangles $ EAD$ and $ EBC$, respectively. Prove that the points $ F$, $ H_{1}$, $ H_{2}$ are collinear.

Original formulation:

Let $ ABC$ be a triangle. A circle passing through $ B$ and $ C$ intersects the sides $ AB$ and $ AC$ again at $ C'$ and $ B',$ respectively. Prove that $ BB'$, $CC'$ and $ HH'$ are concurrent, where $ H$ and $ H'$ are the orthocentres of triangles $ ABC$ and $ AB'C'$ respectively.
40 replies
Stun
Mar 13, 2005
mathwiz_1207
an hour ago
JBMO TST Bosnia and Herzegovina 2023 P3
FishkoBiH   1
N 2 hours ago by Maths_VC
Source: JBMO TST Bosnia and Herzegovina 2023 P3
Let ABC be an acute triangle with an incenter $I$.The Incircle touches sides $AC$ and $AB$ in $E$ and $F$ ,respectively. Lines CI and EF intersect at $S$. The point $T$$I$ is on the line AI so that $EI$=$ET$.If $K$ is the foot of the altitude from $C$ in triangle $ABC$,prove that points $K$,$S$ and $T$ are colinear.
1 reply
FishkoBiH
4 hours ago
Maths_VC
2 hours ago
Rational sequences
tenniskidperson3   58
N 2 hours ago by meduh6849
Source: 2009 USAMO problem 6
Let $s_1, s_2, s_3, \dots$ be an infinite, nonconstant sequence of rational numbers, meaning it is not the case that $s_1 = s_2 = s_3 = \dots.$ Suppose that $t_1, t_2, t_3, \dots$ is also an infinite, nonconstant sequence of rational numbers with the property that $(s_i - s_j)(t_i - t_j)$ is an integer for all $i$ and $j$. Prove that there exists a rational number $r$ such that $(s_i - s_j)r$ and $(t_i - t_j)/r$ are integers for all $i$ and $j$.
58 replies
tenniskidperson3
Apr 30, 2009
meduh6849
2 hours ago
number theory diophantic with factorials and primes
skellyrah   5
N 2 hours ago by GreekIdiot
Source: by me
find all triplets of non negative integers (a,b,p) where p is prime such that $$ a! + b! + 7ab = p^2 $$
5 replies
skellyrah
Feb 16, 2025
GreekIdiot
2 hours ago
Rational coefficients polynomial
Cats_on_a_computer   0
2 hours ago
Given a quartic monic polynomial with rational coefficients, show that if the polynomial has exactly 1 real root r, r must be rational.
I solved this somewhat differently (using the division algorithm), but it really seems like Vieta should work here. I haven’t been able to find another workable solution however.
0 replies
Cats_on_a_computer
2 hours ago
0 replies
JBMO TST Bosnia and Herzegovina 2024 P2
FishkoBiH   1
N 2 hours ago by Rotten_
Source: JBMO TST Bosnia and Herzegovina 2024 P2
Determine all $x$, $y$, $k$ and $n$ positive integers such that:

$10^x$ + $10^y$ + $n!$ = $2024^k$

1 reply
FishkoBiH
3 hours ago
Rotten_
2 hours ago
4th grader qual JMO
HCM2001   38
N 4 hours ago by blueprimes
i mean.. whattttt??? just found out about this.. is he on aops? (i'm sure he is) where are you orz lol..
https://www.mathschool.com/blog/results/celebrating-success-douglas-zhang-is-rsm-s-youngest-usajmo-qualifier
38 replies
HCM2001
May 22, 2025
blueprimes
4 hours ago
basic nt
zhoujef000   39
N Apr 20, 2025 by NicoN9
Source: 2025 AIME I #1
Find the sum of all integer bases $b>9$ for which $17_b$ is a divisor of $97_b.$
39 replies
zhoujef000
Feb 7, 2025
NicoN9
Apr 20, 2025
basic nt
G H J
G H BBookmark kLocked kLocked NReply
Source: 2025 AIME I #1
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zhoujef000
322 posts
#1 • 2 Y
Y by PikaPika999, Soupboy0
Find the sum of all integer bases $b>9$ for which $17_b$ is a divisor of $97_b.$
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chessboy_123
167 posts
#2 • 1 Y
Y by PikaPika999
21 and 49 --> 070?
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plang2008
337 posts
#3 • 1 Y
Y by PikaPika999
$b + 7 \mid 9b + 7 \implies b + 7 \mid -56$ so $b = 21, 49 \implies \boxed{070}$.
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QuelFromage
14 posts
#4
Y by
$9b+7 \equiv 0~(\text{mod }b+7) \implies 9b+7 - 9(b+7) \equiv 0~(\text{mod } b+7) \implies -56 \equiv 0~(\text{mod } b+7) \implies b = 21, 49 \implies \boxed{070}$
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MathPerson12321
3795 posts
#5
Y by
QuelFromage wrote:
$9b+7 \equiv 0~(\text{mod }b+7) \implies 9b+7 - 9(b+7) \equiv 0~(\text{mod } b+7) \implies -56 \equiv 0~(\text{mod } b+7) \implies b = 21, 49 \implies \boxed{070}$

I forgot how to modulo so I spent a ton of time looking for other sols until I realized it was just 70.
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ChaitraliKA
1015 posts
#6
Y by
I did Simon's factoring trick lol
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fruitmonster97
2502 posts
#7
Y by
$b+7|9b+7$ so $b+7|8b$ so let $k(b+7)=8b$ and check $k<8$ to find $21+49=\boxed{070}$
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xHypotenuse
787 posts
#8
Y by
Let a = b+7 and then you find that b+7 | 56 so only b = 21 and b= 49 work when b>9. Therefore, 21 + 40 = 070.
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darkgrimoire
5 posts
#9
Y by
Did anyone use polynomial division to get 9-\frac{56}{b+7}. Then b+7=28, and b+7 =56. I forgot all the rules of divisibility lol.
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BS2012
1052 posts
#10 • 1 Y
Y by darkgrimoire
darkgrimoire wrote:
Did anyone use polynomial division to get 9-\frac{56}{b+7}. Then b+7=28, and b+7 =56. I forgot all the rules of divisibility lol.

yea thats what i did
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darkgrimoire
5 posts
#11
Y by
darkgrimoire wrote:
Did anyone use polynomial division to get $9-\frac{56}{b+7}$. Then b+7=28, and b+7 =56. I forgot all the rules of divisibility lol.
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DreamineYT
298 posts
#12
Y by
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megahertz13
3191 posts
#13
Y by
Notice that we have $$7b+9\equiv 0\pmod {b+9}\implies 56\equiv 0\pmod {b+9}.$$Now by inspection, $b=21$ and $b=49$ are the only solutions. This yields the answer $\boxed{070}$.
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sixoneeight
1138 posts
#14
Y by
Easy things give $b+7|56$. Thus, the possible values greater than $9$ are $14, 28, 56$.

1 hour later

Oh wait, forgot to subtract $7$ hahaha
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megarnie
5610 posts
#15 • 1 Y
Y by Sedro
Fakesolve that gives right answer:

$b + 7 \mid 9b + 7\implies b + 7 \mid 9b$ (this is wrong but it's what i did), so $b + 7 \mid 63$. Thus, $b \in \{14,56\}$, so the answer is $\boxed{070}$.
This post has been edited 3 times. Last edited by megarnie, Feb 7, 2025, 5:52 PM
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MathRook7817
750 posts
#16
Y by
megarnie wrote:
Fakesolve that gives right answer:

$b + 7 \mid 9b + 7\implies b + 7 \mid 9b$ (this is wrong but it's what i did), so $b + 7 \mid 63$. Thus, $b \in \{14,56\}$, so the answer is $\boxed{070}$.

so lucky lol
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AshAuktober
1009 posts
#17
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$b+7\mid 9b+7\iff b+7 \mid 56$, so $b = 21, 49$.Therefore the answer is $\boxed{70}$.
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theorz
6 posts
#18
Y by
ez it is 70
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pingpongmerrily
3763 posts
#19
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b+7 divides 9b+7
b-49 divides b+7
b+7 divides 56
b=21, b=49
070
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lpieleanu
3004 posts
#20
Y by
Solution
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junlongsun
70 posts
#21
Y by
$\frac{9b+7}{b+7}$ has to be an integer
Notice that we can rewrite $9b+7$ as $9(b+7)-56$
$$\frac{9b+7}{b+7}=\frac{9(b+7)-56}{b+7}=\frac{9(b+7)}{b+7}+\frac{-56}{b+7}$$Because $\frac{9(b+7)}{b+7}=9$, we just need to make sure $\frac{-56}{b+7}$ is an integer.

The only factors of 56 that satisfy $b>9$, $b+7=28$, $b+7=56$
$$b=21, b=49$$$$49+21=70$$$$\fbox{70}$$
This post has been edited 3 times. Last edited by junlongsun, Feb 8, 2025, 12:05 AM
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apple143
62 posts
#22
Y by
AshAuktober wrote:
$b+7\mid 9b+7\iff b+7 \mid 56$, so $b = 21, 49$.Therefore the answer is $\boxed{70}$.

yeah this is what i got. wrote almost nothing down
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ashays
91 posts
#23
Y by
9b+7 is divisible by b+7, so 9b+7 - 9(b+7) will still be divisible by b+7, so -56 is divisible by b+7. we first try -56-7=-49, so one b possibility could be 49. 49-7=42, 42/2 = 21, 21 is another possibility. 21-7=14, 14/2 = 7, which is less than 9, so we only have 21 and 49, giving us 70.
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RedFireTruck
4243 posts
#24
Y by
We want $(b+7)|(9b+7)$ so $(b+7)|56$ so the answer is $(56-7)+(28-7)=49+21=\boxed{070}$.
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gracemoon124
872 posts
#25
Y by
$17_b$ dividing $97_b$ means that $b+7\mid 9b+7$, and note that $9b+7=9(b+7)-56$, so $b+7\mid 56$. Then $b$ could be $21$ or $49$ which means $\boxed{070}$.

remarks: while mocking this i had a worse solution (bounding based on value of $\tfrac{9b+7}{b+7}$ WHICH WORKS THOUGH) and it takes about the same amount of time (:
This post has been edited 1 time. Last edited by gracemoon124, Feb 8, 2025, 6:17 PM
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pog
4917 posts
#26
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if you do not know any number theory:

By the definition of a divisor, $(b+7)k = 9b+7$ for an integer $k$. Clearly $k \ge 9$ does not work. So now we can just try every single value.


$b+7=9b+7$ no
$2b+14=9b+7$ gives $7b=7$ which doesn't work
$3b+21=9b+7$ gives $6b=14$ which doesn't work
$4b+28=9b+7$ gives $5b=21$ which doesn't work
$5b+35=9b+7$ gives $4b=28$ which doesn't work
$6b+42=9b+7$ gives $3b=35$ which doesn't work
$7b+49=9b+7$ gives $2b=42$ which does work
$8b+56=9b+7$ gives $1b=49$ which does work


Hence our answer is $21+49=\boxed{70}$.
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Mr.Sharkman
501 posts
#27
Y by
LMFAO this is what I thought p1 was gonna be: What is the sum of all $n$ for which $15_{n}$ divides $2025_{n}$?
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pingpongmerrily
3763 posts
#28
Y by
Mr.Sharkman wrote:
LMFAO this is what I thought p1 was gonna be: What is the sum of all $n$ for which $15_{n}$ divides $2025_{n}$?

that might be hard
uh (n+5) divides $2n^3+2n+5$
is that even factorable
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sunannie
13 posts
#29
Y by
pog wrote:
if you do not know any number theory:

By the definition of a divisor, $(b+7)k = 9b+7$ for an integer $k$. Clearly $k \ge 9$ does not work. So now we can just try every single value.


$b+7=9b+7$ no
$2b+14=9b+7$ gives $7b=7$ which doesn't work
$3b+21=9b+7$ gives $6b=14$ which doesn't work
$4b+28=9b+7$ gives $5b=21$ which doesn't work
$5b+35=9b+7$ gives $4b=28$ which doesn't work
$6b+42=9b+7$ gives $3b=35$ which doesn't work
$7b+49=9b+7$ gives $2b=42$ which does work
$8b+56=9b+7$ gives $1b=49$ which does work


Hence our answer is $21+49=\boxed{70}$.

thats what i did too
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xTimmyG
265 posts
#30 • 1 Y
Y by MrMustache
by the fundamental theory of number theory, the answer must start with an even number. then, guess and check values of b, to get 21+49=70
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MrMustache
3071 posts
#31
Y by
+1 for guess and check its really the only good way to solve this problem. I did all values up to 187 just to be sure.
This post has been edited 1 time. Last edited by MrMustache, Feb 8, 2025, 10:13 PM
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aidan0626
1936 posts
#32
Y by
Mr.Sharkman wrote:
LMFAO this is what I thought p1 was gonna be: What is the sum of all $n$ for which $15_{n}$ divides $2025_{n}$?

that might be hard
uh (n+5) divides $2n^3+2n+5$
is that even factorable

that wouldn't be that much harder
you get $n+5|255$, and it's basically the same thing
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sansgankrsngupta
147 posts
#34
Y by
OG!
$$(17)_b \mid 97_b \iff b^2+7b \mid 9b^2+7b \iff b+7 \mid 8b \iff b+7 \mid 56$$.
Since $b>9$, the only possible $b$ are $b=21,49$ which satisfy.
Thus, the sum of all possible values of $b = 21+49= \boxed{070}$
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sadas123
1318 posts
#35
Y by
xHypotenuse wrote:
Let a = b+7 and then you find that b+7 | 56 so only b = 21 and b= 49 work when b>9. Therefore, 21 + 40 = 070.

what 21+40=61 I think you meant 49??
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sadas123
1318 posts
#36
Y by
I was stuck on this problem for some reason I don't know why

The best way to tackle this problem is to convert everything to variables so first 17 base b = b + 7 and 97 base b = 9b+7 so we can just make a variable when multiplied by it it equals 97 base b

So first to do that we can simplify it like 9b+7 = x(b+7) for some value of x then when we multiply that out we get 9b+7=bx+7x. Because we want to solve for xb we have to subtract that on both sides to get 9b+7-bx= 7x then in any situation like this we have to factor out the b, but first we can subtract 7 on both sides.

When we do that we get b(9-x)= 7(x-1) which when we divide both sides we get

b= (7(x-1))/(9-x) we get this easy equation to solve because we know that x has to be a single digit number because anything greater than 9 won't work so we get that after guess and check x= 7 and 8 so when we plug that in we get 21 and 49, so when we add those together you get $49+21$ $=$ $70$
This post has been edited 2 times. Last edited by sadas123, Feb 9, 2025, 4:44 PM
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Mr.Sharkman
501 posts
#37
Y by
pingpongmerrily wrote:
Mr.Sharkman wrote:
LMFAO this is what I thought p1 was gonna be: What is the sum of all $n$ for which $15_{n}$ divides $2025_{n}$?

that might be hard
uh (n+5) divides $2n^3+2n+5$
is that even factorable

Bruh factor theorem
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A04572
9 posts
#38
Y by
b+7 is divisible by 9b+7, but because b+7 is divisible by 9b+63, then b+7 is divisible by 56. And B+7 is greater than 7, so b=21 or 49[Click][sounds stupid but I got b+7 is divisible by 8b, and did a lot of stupid stuff to get -6+-5+-3+0+1+7+21+49=64 cause I forgot b is greater than 9, but luckily I realized this at the end.]
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jasperE3
11383 posts
#39
Y by
zhoujef000 wrote:
Find the sum of all integer bases $b>9$ for which $17_b$ is a divisor of $97_b.$

We need $b+7\mid9b+7$, so $b+7\mid8b$. Let $8b=k(b+7)$, do casework and bounding.
If $k\le4$ then $8b\le4(b+7)$ so $b\le7$, doesn't work.
If $k=5$ then $8b=5b+35$ so $b=\frac{35}3\notin\mathbb Z$, doesn't work.
If $k=6$ then $8b=6b+42$ so $b=21$, which does work.
If $k=7$ then $8b=7b+49$ so $b=49$, which does work.
If $k\ge8$ then $8b\ge8(b+7)$ so $0\ge56$, doesn't work.
Our answer is $21+49=\boxed{070}$.
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Apple_maths60
27 posts
#40
Y by
(97)b =9b+7
(17)b=b+7
So b+7 divides 9b+7
gcd(9b+7,b+7)=b+7
gcd(-56,b+7)=b+7
So we get b+7 divides -56
Now , 56=2*2*2*7 and factors greater than 9+7 are 28 and 56 itself
So, b=21 or 49
Sum =70(answer)
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NicoN9
157 posts
#41
Y by
$b+7\mid 9b+7 \Longleftrightarrow b+7\mid -56$ and we easily get $b=21, 49$.
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