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zhoujef000

322 posts

zhoujef000

#1 h Feb 7, 2025, 3:46 PM • 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

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chessboy_123

#2 h Feb 7, 2025, 3:47 PM • 1 Y

Y by PikaPika999

21 and 49 --> 070?

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plang2008

337 posts

plang2008

#3 h Feb 7, 2025, 3:47 PM • 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

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QuelFromage

#4 h Feb 7, 2025, 4:02 PM

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$9b+7 \equiv 0~(\text{mod }b+7) \implies 9b+7 - 9(b+7) \equiv 0~(\text{mod } b+7) \implies -56 \equiv 0~(\text{mod } b+7) \implies b = 21, 49 \implies \boxed{070}$

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MathPerson12321

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MathPerson12321

#5 h Feb 7, 2025, 4:08 PM

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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

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ChaitraliKA

#6 h Feb 7, 2025, 4:11 PM

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I did Simon's factoring trick lol

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fruitmonster97

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fruitmonster97

#7 h Feb 7, 2025, 4:14 PM

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$b+7|9b+7$ 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

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xHypotenuse

#8 h Feb 7, 2025, 4:59 PM

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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

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darkgrimoire

#9 h Feb 7, 2025, 5:02 PM

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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

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BS2012

#10 h Feb 7, 2025, 5:03 PM • 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

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darkgrimoire

#11 h Feb 7, 2025, 5:04 PM

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

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DreamineYT

#12 h Feb 7, 2025, 5:12 PM

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Click to reveal hidden text

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megahertz13

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megahertz13

#13 h Feb 7, 2025, 5:28 PM

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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

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sixoneeight

#14 h Feb 7, 2025, 5:37 PM

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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

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megarnie

#15 h Feb 7, 2025, 5:51 PM • 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

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MathRook7817

#16 h Feb 7, 2025, 6:01 PM

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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

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AshAuktober

#17 h Feb 7, 2025, 6:16 PM

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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

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theorz

#18 h Feb 7, 2025, 6:24 PM

Y by

ez it is 70

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pingpongmerrily

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pingpongmerrily

#19 h Feb 7, 2025, 6:37 PM

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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

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lpieleanu

#20 h Feb 7, 2025, 7:51 PM

Y by

Solution

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junlongsun

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junlongsun

#21 h Feb 7, 2025, 9:00 PM

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$\frac{9b+7}{b+7}$ 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

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apple143

#22 h Feb 8, 2025, 1:50 AM

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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

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ashays

#23 h Feb 8, 2025, 2:04 AM

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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

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RedFireTruck

#24 h Feb 8, 2025, 4:19 PM

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

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gracemoon124

#25 h Feb 8, 2025, 6:17 PM

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

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pog

#26 h Feb 8, 2025, 6:21 PM

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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

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Mr.Sharkman

#27 h Feb 8, 2025, 9:58 PM

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

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pingpongmerrily

#28 h Feb 8, 2025, 10:00 PM

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

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sunannie

#29 h Feb 8, 2025, 10:04 PM

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

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xTimmyG

#30 h Feb 8, 2025, 10:07 PM • 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

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MrMustache

#31 h Feb 8, 2025, 10:12 PM

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

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aidan0626

#32 h Feb 8, 2025, 10:51 PM

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

that wouldn't be that much harder
you get $n+5|255$ , and it's basically the same thing

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sansgankrsngupta

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sansgankrsngupta

#34 h Feb 9, 2025, 2:53 PM

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

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sadas123

#35 h Feb 9, 2025, 3:00 PM

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

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sadas123

#36 h Feb 9, 2025, 4:43 PM

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

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Mr.Sharkman

#37 h Feb 11, 2025, 1:30 PM

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

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A04572

#38 h Feb 22, 2025, 8:48 AM

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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

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jasperE3

#39 h Feb 28, 2025, 4:49 AM

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

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Apple_maths60

#40 h Apr 5, 2025, 6:08 PM

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

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NicoN9

#41 h Apr 20, 2025, 6:25 AM

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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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