Stay ahead of learning milestones! Enroll in a class over the summer!

Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
3 M G
BBookmark  VNew Topic kLocked
Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
3 M G
BBookmark  VNew Topic kLocked
G
Topic
First Poster
Last Poster
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.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
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 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
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 Self-Paced

Introduction to Counting & Probability
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
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 Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
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

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
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

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
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
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
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

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
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

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
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
Jane street swag package? USA(J)MO
arfekete   8
N 3 minutes ago by elasticwealth
Hey! People are starting to get their swag packages from Jane Street for qualifying for USA(J)MO, and after some initial discussion on what we got, people are getting different things. Out of curiosity, I was wondering how they decide who gets what.
Please enter the following info:

- USAMO or USAJMO
- Grade
- Score
- Award/Medal/HM
- MOP (yes or no, if yes then color)
- List of items you got in your package

I will reply with my info as an example.
8 replies
+1 w
arfekete
Yesterday at 4:34 PM
elasticwealth
3 minutes ago
USAMO Medals
YauYauFilter   45
N 5 minutes ago by mhgelgi
+1 w
YauYauFilter
Apr 24, 2025
mhgelgi
5 minutes ago
9 ARML Location
deduck   36
N 9 minutes ago by idk12345678
UNR -> Nevada
St Anselm -> New Hampshire
PSU -> Pennsylvania
WCU -> North Carolina


Put your USERNAME in the list ONLY IF YOU WANT TO!!!! !!!!!

I'm going to UNR if anyone wants to meetup!!! :D

Current List:
Iowa
UNR
PSU
St Anselm
WCU
36 replies
+1 w
deduck
Tuesday at 4:19 PM
idk12345678
9 minutes ago
A problem from Le Anh Vinh book.
minhquannguyen   0
28 minutes ago
Source: LE ANH VINH, DINH HUONG BOI DUONG HOC SINH NANG KHIEU TOAN TAP 1 DAI SO
Let $n$ is a positive integer. Determine all functions $f:(1,+\infty)\to\mathbb{R}$ such that
\[f(x^{n+1}+y^{n+1})=x^nf(x)+y^nf(y),\forall x,y>1.\]
0 replies
minhquannguyen
28 minutes ago
0 replies
IMO ShortList 1999, algebra problem 1
orl   42
N an hour ago by ihategeo_1969
Source: IMO ShortList 1999, algebra problem 1
Let $n \geq 2$ be a fixed integer. Find the least constant $C$ such the inequality

\[\sum_{i<j} x_{i}x_{j} \left(x^{2}_{i}+x^{2}_{j} \right) \leq C
\left(\sum_{i}x_{i} \right)^4\]

holds for any $x_{1}, \ldots ,x_{n} \geq 0$ (the sum on the left consists of $\binom{n}{2}$ summands). For this constant $C$, characterize the instances of equality.
42 replies
orl
Nov 13, 2004
ihategeo_1969
an hour ago
how prestigious is hsmc
ConfidentKoala4   3
N an hour ago by ConfidentKoala4
been wonderin this for a while

how prestigious is it? ik its not as good as mathily (they rejected me :mad: ) but Idk how good it actually is
3 replies
ConfidentKoala4
2 hours ago
ConfidentKoala4
an hour ago
q(x) to be the product of all primes less than p(x)
orl   19
N an hour ago by ihategeo_1969
Source: IMO Shortlist 1995, S3
For an integer $x \geq 1$, let $p(x)$ be the least prime that does not divide $x$, and define $q(x)$ to be the product of all primes less than $p(x)$. In particular, $p(1) = 2.$ For $x$ having $p(x) = 2$, define $q(x) = 1$. Consider the sequence $x_0, x_1, x_2, \ldots$ defined by $x_0 = 1$ and \[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \] for $n \geq 0$. Find all $n$ such that $x_n = 1995$.
19 replies
orl
Aug 10, 2008
ihategeo_1969
an hour ago
Interesting inequality
sealight2107   2
N 2 hours ago by arqady
Source: Own
Let $a,b,c>0$ such that $a+b+c=3$. Find the minimum value of:
$Q=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}$
2 replies
sealight2107
Tuesday at 4:53 PM
arqady
2 hours ago
Cyclic Quads and Parallel Lines
gracemoon124   16
N 3 hours ago by ohiorizzler1434
Source: 2015 British Mathematical Olympiad?
Let $ABCD$ be a cyclic quadrilateral. Let $F$ be the midpoint of the arc $AB$ of its circumcircle which does not contain $C$ or $D$. Let the lines $DF$ and $AC$ meet at $P$ and the lines $CF$ and $BD$ meet at $Q$. Prove that the lines $PQ$ and $AB$ are parallel.
16 replies
gracemoon124
Aug 16, 2023
ohiorizzler1434
3 hours ago
Radical Center on the Euler Line (USEMO 2020/3)
franzliszt   37
N 4 hours ago by Ilikeminecraft
Source: USEMO 2020/3
Let $ABC$ be an acute triangle with circumcenter $O$ and orthocenter $H$. Let $\Gamma$ denote the circumcircle of triangle $ABC$, and $N$ the midpoint of $OH$. The tangents to $\Gamma$ at $B$ and $C$, and the line through $H$ perpendicular to line $AN$, determine a triangle whose circumcircle we denote by $\omega_A$. Define $\omega_B$ and $\omega_C$ similarly.
Prove that the common chords of $\omega_A$,$\omega_B$ and $\omega_C$ are concurrent on line $OH$.

Proposed by Anant Mudgal
37 replies
franzliszt
Oct 24, 2020
Ilikeminecraft
4 hours ago
Functional equation with powers
tapir1729   13
N 4 hours ago by ihategeo_1969
Source: TSTST 2024, problem 6
Determine whether there exists a function $f: \mathbb{Z}_{> 0} \rightarrow \mathbb{Z}_{> 0}$ such that for all positive integers $m$ and $n$,
\[f(m+nf(m))=f(n)^m+2024! \cdot m.\]Jaedon Whyte
13 replies
tapir1729
Jun 24, 2024
ihategeo_1969
4 hours ago
Powers of a Prime
numbertheorist17   34
N 4 hours ago by KevinYang2.71
Source: USA TSTST 2014, Problem 6
Suppose we have distinct positive integers $a, b, c, d$, and an odd prime $p$ not dividing any of them, and an integer $M$ such that if one considers the infinite sequence \begin{align*}
		ca &- db \\
		ca^2 &- db^2 \\
		ca^3 &- db^3 \\
		ca^4 &- db^4 \\
&\vdots
	\end{align*} and looks at the highest power of $p$ that divides each of them, these powers are not all zero, and are all at most $M$. Prove that there exists some $T$ (which may depend on $a,b,c,d,p,M$) such that whenever $p$ divides an element of this sequence, the maximum power of $p$ that divides that element is exactly $p^T$.
34 replies
numbertheorist17
Jul 16, 2014
KevinYang2.71
4 hours ago
IMO 2018 Problem 5
orthocentre   80
N 5 hours ago by OronSH
Source: IMO 2018
Let $a_1$, $a_2$, $\ldots$ be an infinite sequence of positive integers. Suppose that there is an integer $N > 1$ such that, for each $n \geq N$, the number
$$\frac{a_1}{a_2} + \frac{a_2}{a_3} + \cdots + \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1}$$is an integer. Prove that there is a positive integer $M$ such that $a_m = a_{m+1}$ for all $m \geq M$.

Proposed by Bayarmagnai Gombodorj, Mongolia
80 replies
orthocentre
Jul 10, 2018
OronSH
5 hours ago
Line passes through fixed point, as point varies
Jalil_Huseynov   60
N 5 hours ago by Rayvhs
Source: APMO 2022 P2
Let $ABC$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $CB$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $AD$ and let $F$ be the seconf intersection point of the circumcircle of $\triangle ACD$ and the circumcircle of $\triangle BDE$. Prove that as $D$ varies, the line $EF$ passes through a fixed point.
60 replies
Jalil_Huseynov
May 17, 2022
Rayvhs
5 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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
zhoujef000
316 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 $97_b.$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
chessboy_123
168 posts
#2 • 1 Y
Y by PikaPika999
21 and 49 --> 070?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
plang2008
337 posts
#3 • 1 Y
Y by PikaPika999
$b + 7 \mid 9b + 7 \implies b + 7 \mid -56$ so $b = 21, 49 \implies \boxed{070}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
QuelFromage
14 posts
#4
Y by
$9b+7 \equiv 0~(\text{mod }b+7) \implies 9b+7 - 9(b+7) \equiv 0~(\text{mod } b+7) \implies -56 \equiv 0~(\text{mod } b+7) \implies b = 21, 49 \implies \boxed{070}$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MathPerson12321
3747 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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ChaitraliKA
1004 posts
#6
Y by
I did Simon's factoring trick lol
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
fruitmonster97
2492 posts
#7
Y by
$b+7|9b+7$ so $b+7|8b$ so let $k(b+7)=8b$ and check $k<8$ to find $21+49=\boxed{070}$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
xHypotenuse
778 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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
BS2012
1033 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
DreamineYT
287 posts
#12
Y by
Click to reveal hidden text
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
megahertz13
3183 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 $\boxed{070}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
megarnie
5606 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MathRook7817
683 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AshAuktober
1005 posts
#17
Y by
$b+7\mid 9b+7\iff b+7 \mid 56$, so $b = 21, 49$.Therefore the answer is $\boxed{70}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
theorz
6 posts
#18
Y by
ez it is 70
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pingpongmerrily
3615 posts
#19
Y by
b+7 divides 9b+7
b-49 divides b+7
b+7 divides 56
b=21, b=49
070
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lpieleanu
3001 posts
#20
Y by
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
RedFireTruck
4223 posts
#24
Y by
We want $(b+7)|(9b+7)$ so $(b+7)|56$ so the answer is $(56-7)+(28-7)=49+21=\boxed{070}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pog
4917 posts
#26
Y by
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 $21+49=\boxed{70}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mr.Sharkman
500 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 $2025_{n}$?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pingpongmerrily
3615 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MrMustache
3015 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
aidan0626
1898 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sansgankrsngupta
143 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 $b = 21+49= \boxed{070}$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sadas123
1265 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??
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sadas123
1265 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mr.Sharkman
500 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.]
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jasperE3
11304 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 $21+49=\boxed{070}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Apple_maths60
26 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)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
NicoN9
147 posts
#41
Y by
$b+7\mid 9b+7 \Longleftrightarrow b+7\mid -56$ and we easily get $b=21, 49$.
Z K Y
N Quick Reply
G
H
=
a