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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)!

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[*]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]
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0 replies
jlacosta
May 1, 2025
0 replies
9 ARML Location
deduck   22
N 14 minutes ago by EaZ_Shadow
UNR -> Nevada
St Anselm -> New Hampshire
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
22 replies
deduck
Today at 4:19 PM
EaZ_Shadow
14 minutes ago
SUMaC Residential vs. Ross
AwesomeDude10   3
N 37 minutes ago by miguel00
Hi! I got into the SUMaC residential i program, and I also recently got off the Ross waitlist. I was wondering if anyone had any insight on which program is
1) More useful in furthering my mathematical knowledge (which has a better curriculum)
2) Since I'm a junior, which is more useful for college apps? (I know this is a little cringe)
Thanks!
3 replies
AwesomeDude10
an hour ago
miguel00
37 minutes ago
I'm trying to find a good math comp...
ysn613   9
N 2 hours ago by anticodon
Okay, so I'm in sixth grade. I have been doing AMC 8 since fourth grade, but not anything else. I was wondering what other "good" math competitions there are that I am the right age for.

I'm also looking for prep tips for math competitions, because when I (mock)ace 2000-2010 AMC 8 and then get a 19 on the real thing when I was definitely able to solve everything, I feel like what I'm doing isn't really working. Anyone got any ideas? Thanks!
9 replies
ysn613
Apr 30, 2025
anticodon
2 hours ago
USAMO MERCH
elasticwealth   5
N 5 hours ago by ninjaforce
Jane street sent:
- A T-shirt
- Deck of cards
- Portable charger (battery)
- 12 oz mug
- Jane Street Hat
5 replies
elasticwealth
Today at 2:28 PM
ninjaforce
5 hours ago
Putnam 2016 A1
Kent Merryfield   15
N Today at 10:51 AM by anudeep
Find the smallest positive integer $j$ such that for every polynomial $p(x)$ with integer coefficients and for every integer $k,$ the integer
\[p^{(j)}(k)=\left. \frac{d^j}{dx^j}p(x) \right|_{x=k}\](the $j$-th derivative of $p(x)$ at $k$) is divisible by $2016.$
15 replies
Kent Merryfield
Dec 4, 2016
anudeep
Today at 10:51 AM
Determinant is 1
Entrepreneur   2
N Today at 8:27 AM by Entrepreneur
If a determinant is of $n^{\text{th}}$ order, and if the constituents of its first, second, ..., $n^{\text{th}}$ rows are the first $n$ figurate numbers of the first, second, ..., $n^{\text{th}}$ orders respectively, show that it's value is $1.$
2 replies
Entrepreneur
Yesterday at 7:14 PM
Entrepreneur
Today at 8:27 AM
Can cos(√2 t) be expressed as a polynomial in cost?
tom-nowy   1
N Today at 7:17 AM by Aiden-1089
Source: Question arising while viewing https://artofproblemsolving.com/community/c51293h3562250
Can $\cos ( \sqrt{2}\,  t )$ be expressed as a polynomial in $\cos t$ with real coefficients?
1 reply
tom-nowy
Today at 7:10 AM
Aiden-1089
Today at 7:17 AM
36x⁴ + 12x² - 36x + 13 > 0
fxandi   2
N Today at 7:14 AM by MeKnowsNothing
Prove that for any real $x \geq 0$ holds inequality $36x^4 + 12x^2 - 36x + 13 > 0.$
2 replies
fxandi
Yesterday at 10:02 PM
MeKnowsNothing
Today at 7:14 AM
2024 Putnam A1
KevinYang2.71   20
N Today at 5:50 AM by thelateone
Determine all positive integers $n$ for which there exists positive integers $a$, $b$, and $c$ satisfying
\[
2a^n+3b^n=4c^n.
\]
20 replies
KevinYang2.71
Dec 10, 2024
thelateone
Today at 5:50 AM
2025 OMOUS Problem 4
enter16180   2
N Yesterday at 8:57 PM by Acridian9
Source: Open Mathematical Olympiad for University Students (OMOUS-2025)
Find all matrices $M \in M_{n}(\mathbb{C})$ such that following equality holds

$$
\operatorname{rank}(M)+\operatorname{rank}\left(M^{2023}-M^{2025}\right)=\operatorname{rank}\left(M-M^{2}\right)+\operatorname{rank}\left(M^{2023}+M^{2024}\right)
$$
2 replies
enter16180
Apr 18, 2025
Acridian9
Yesterday at 8:57 PM
Poker hand
Aksudon   1
N Yesterday at 6:32 PM by lucaminiati
Problem: In a standard 52-card deck, how many different five-card poker hands are there of 'two pairs'?

Can someone please explain what is logically wrong with the following solution? (It gives double of the right solution which supposed to be 123552).

13\binom{4}{2}*12\binom{4}{2}*44=247104

Thanks
1 reply
Aksudon
Yesterday at 5:14 PM
lucaminiati
Yesterday at 6:32 PM
Sequence with GCD involved
mathematics2004   3
N Yesterday at 5:54 PM by anudeep
Source: 2021 Simon Marais, A2
Define the sequence of integers $a_1, a_2, a_3, \ldots$ by $a_1 = 1$, and
\[ a_{n+1} = \left(n+1-\gcd(a_n,n) \right) \times a_n \]for all integers $n \ge 1$.
Prove that $\frac{a_{n+1}}{a_n}=n$ if and only if $n$ is prime or $n=1$.
Here $\gcd(s,t)$ denotes the greatest common divisor of $s$ and $t$.
3 replies
mathematics2004
Nov 2, 2021
anudeep
Yesterday at 5:54 PM
Putnam 2000 B2
ahaanomegas   20
N Yesterday at 5:05 PM by reni_wee
Prove that the expression \[ \dfrac {\text {gcd}(m, n)}{n} \dbinom {n}{m} \] is an integer for all pairs of integers $ n \ge m \ge 1 $.
20 replies
ahaanomegas
Sep 6, 2011
reni_wee
Yesterday at 5:05 PM
f(x)f'(x)≥cos, f(∞)=undef. if f is bounded
jasperE3   2
N Yesterday at 4:21 PM by Rohit-2006
Source: VJIMC 2013 1.1
Let $f:[0,\infty)\to\mathbb R$ be a differentiable function with $|f(x)|\le M$ and $f(x)f'(x)\ge\cos x$ for $x\in[0,\infty)$, where $M>0$. Prove that $f(x)$ does not have a limit as $x\to\infty$.
2 replies
jasperE3
May 30, 2021
Rohit-2006
Yesterday at 4:21 PM
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
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 $97_b.$
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chessboy_123
168 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
3744 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
1004 posts
#6
Y by
I did Simon's factoring trick lol
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fruitmonster97
2491 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
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.
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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
1029 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
287 posts
#12
Y by
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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 $\boxed{070}$.
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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
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
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MathRook7817
674 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
1005 posts
#17
Y by
$b+7\mid 9b+7\iff b+7 \mid 56$, so $b = 21, 49$.Therefore the answer is $\boxed{70}$.
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theorz
6 posts
#18
Y by
ez it is 70
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pingpongmerrily
3606 posts
#19
Y by
b+7 divides 9b+7
b-49 divides b+7
b+7 divides 56
b=21, b=49
070
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lpieleanu
2994 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
4223 posts
#24
Y by
We want $(b+7)|(9b+7)$ so $(b+7)|56$ so the answer is $(56-7)+(28-7)=49+21=\boxed{070}$.
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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
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 $21+49=\boxed{70}$.
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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 $2025_{n}$?
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pingpongmerrily
3606 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
3013 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
1897 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
144 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
1262 posts
#35
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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
1262 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
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
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A04572
9 posts
#38
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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
11296 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
26 posts
#40
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(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
147 posts
#41
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$b+7\mid 9b+7 \Longleftrightarrow b+7\mid -56$ and we easily get $b=21, 49$.
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