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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 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]
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0 replies
jlacosta
May 1, 2025
0 replies
MAN IS KID
DrMath   136
N 3 hours ago by lakshya2009
Source: USAMO 2017 P3, Evan Chen
Let $ABC$ be a scalene triangle with circumcircle $\Omega$ and incenter $I$. Ray $AI$ meets $\overline{BC}$ at $D$ and meets $\Omega$ again at $M$; the circle with diameter $\overline{DM}$ cuts $\Omega$ again at $K$. Lines $MK$ and $BC$ meet at $S$, and $N$ is the midpoint of $\overline{IS}$. The circumcircles of $\triangle KID$ and $\triangle MAN$ intersect at points $L_1$ and $L_2$. Prove that $\Omega$ passes through the midpoint of either $\overline{IL_1}$ or $\overline{IL_2}$.

Proposed by Evan Chen
136 replies
1 viewing
DrMath
Apr 19, 2017
lakshya2009
3 hours ago
[Sipnayan JHS] Semifinals Round B, Average, #2
LilKirb   1
N 3 hours ago by LilKirb
How many trailing zeroes are there in the base $4$ representation of $2015!$ ?
1 reply
LilKirb
4 hours ago
LilKirb
3 hours ago
2022 SMT Team Round - Stanford Math Tournament
parmenides51   5
N 4 hours ago by vanstraelen
p1. Square $ABCD$ has side length $2$. Let the midpoint of $BC$ be $E$. What is the area of the overlapping region between the circle centered at $E$ with radius $1$ and the circle centered at $D$ with radius $2$? (You may express your answer using inverse trigonometry functions of noncommon values.)


p2. Find the number of times $f(x) = 2$ occurs when $0 \le x \le 2022 \pi$ for the function $f(x) = 2^x(cos(x) + 1)$.


p3. Stanford is building a new dorm for students, and they are looking to offer $2$ room configurations:
$\bullet$ Configuration $A$: a one-room double, which is a square with side length of $x$,
$\bullet$ Configuration $B$: a two-room double, which is two connected rooms, each of them squares with a side length of $y$.
To make things fair for everyone, Stanford wants a one-room double (rooms of configuration $A$) to be exactly $1$ m$^2$ larger than the total area of a two-room double. Find the number of possible pairs of side lengths $(x, y)$, where $x \in N$, $y \in N$, such that $x - y < 2022$.


p4. The island nation of Ur is comprised of $6$ islands. One day, people decide to create island-states as follows. Each island randomly chooses one of the other five islands and builds a bridge between the two islands (it is possible for two bridges to be built between islands $A$ and $B$ if each island chooses the other). Then, all islands connected by bridges together form an island-state. What is the expected number of island-states Ur is divided into?


p5. Let $a, b,$ and $c$ be the roots of the polynomial $x^3 - 3x^2 - 4x + 5$. Compute $\frac{a^4 + b^4}{a + b}+\frac{b^4 + c^4}{b + c}+\frac{c^4 + a^4}{c + a}$.


p6. Carol writes a program that finds all paths on an 10 by 2 grid from cell (1, 1) to cell (10, 2) subject to the conditions that a path does not visit any cell more than once and at each step the path can go up, down, left, or right from the current cell, excluding moves that would make the path leave the grid. What is the total length of all such paths? (The length of a path is the number of cells it passes through, including the starting and ending cells.)


p7. Consider the sequence of integers an defined by $a_1 = 1$, $a_p = p$ for prime $p$ and $a_{mn} = ma_n + na_m$ for $m, n > 1$. Find the smallest $n$ such that $\frac{a_n^2}{2022}$ is a perfect power of $3$.


p8. Let $\vartriangle ABC$ be a triangle whose $A$-excircle, $B$-excircle, and $C$-excircle have radii $R_A$, $R_B$, and $R_C$, respectively. If $R_AR_BR_C = 384$ and the perimeter of $\vartriangle ABC$ is $32$, what is the area of $\vartriangle ABC$?


p9. Consider the set $S$ of functions $f : \{1, 2, . . . , 16\} \to \{1, 2, . . . , 243\}$ satisfying:
(a) $f(1) = 1$
(b) $f(n^2) = n^2f(n)$,
(c) $n |f(n)$,
(d) $f(lcm(m, n))f(gcd(m, n)) = f(m)f(n)$.
If $|S|$ can be written as $p^{\ell_1}_1 \cdot p^{\ell_2}_2 \cdot ... \cdot  p^{\ell_k}_k$ where $p_i$ are distinct primes, compute $p_1\ell_1+p_2\ell_2+. . .+p_k\ell_k$.


p10. You are given that $\log_{10}2 \approx 0.3010$ and that the first (leftmost) two digits of $2^{1000}$ are 10. Compute the number of integers $n$ with $1000 \le n \le 2000$ such that $2^n$ starts with either the digit $8$ or $9$ (in base $10$).


p11. Let $O$ be the circumcenter of $\vartriangle ABC$. Let $M$ be the midpoint of $BC$, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$, respectively, onto the opposite sides. $EF$ intersects $BC$ at $P$. The line passing through $O$ and perpendicular to $BC$ intersects the circumcircle of $\vartriangle ABC$ at $L$ (on the major arc $BC$) and $N$, and intersects $BC$ at $M$. Point $Q$ lies on the line $LA$ such that $OQ$ is perpendicular to $AP$. Given that $\angle BAC = 60^o$ and $\angle AMC = 60^o$, compute $OQ/AP$.


p12. Let $T$ be the isosceles triangle with side lengths $5, 5, 6$. Arpit and Katherine simultaneously choose points $A$ and $K$ within this triangle, and compute $d(A, K)$, the squared distance between the two points. Suppose that Arpit chooses a random point $A$ within $T$ . Katherine plays the (possibly randomized) strategy which given Arpit’s strategy minimizes the expected value of $d(A, K)$. Compute this value.


p13. For a regular polygon $S$ with $n$ sides, let $f(S)$ denote the regular polygon with $2n$ sides such that the vertices of $S$ are the midpoints of every other side of $f(S)$. Let $f^{(k)}(S)$ denote the polygon that results after applying f a total of k times. The area of $\lim_{k \to \infty} f^{(k)}(P)$ where $P$ is a pentagon of side length $1$, can be expressed as $\frac{a+b\sqrt{c}}{d}\pi^m$ for some positive integers $a, b, c, d, m$ where $d$ is not divisible by the square of any prime and $d$ does not share any positive divisors with $a$ and $b$. Find $a + b + c + d + m$.


p14. Consider the function $f(m) = \sum_{n=0}^{\infty}\frac{(n - m)^2}{(2n)!}$ . This function can be expressed in the form $f(m) = \frac{a_m}{e} +\frac{b_m}{4}e$ for sequences of integers $\{a_m\}_{m\ge 1}$, $\{b_m\}_{m\ge 1}$. Determine $\lim_{n \to \infty}\frac{2022b_m}{a_m}$.


p15. In $\vartriangle ABC$, let $G$ be the centroid and let the circumcenters of $\vartriangle BCG$, $\vartriangle CAG$, and $\vartriangle ABG$ be $I, J$, and $K$, respectively. The line passing through $I$ and the midpoint of $BC$ intersects $KJ$ at $Y$. If the radius of circle $K$ is $5$, the radius of circle $J$ is $8$, and $AG = 6$, what is the length of $KY$ ?



PS. You should use hide for answers. Collected here.
5 replies
parmenides51
Jun 30, 2022
vanstraelen
4 hours ago
have you done DCX-Russian?
GoodMorning   83
N Today at 7:30 AM by ray66
Source: 2023 USAJMO Problem 3
Consider an $n$-by-$n$ board of unit squares for some odd positive integer $n$. We say that a collection $C$ of identical dominoes is a maximal grid-aligned configuration on the board if $C$ consists of $(n^2-1)/2$ dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: $C$ then covers all but one square on the board. We are allowed to slide (but not rotate) a domino on the board to cover the uncovered square, resulting in a new maximal grid-aligned configuration with another square uncovered. Let $k(C)$ be the number of distinct maximal grid-aligned configurations obtainable from $C$ by repeatedly sliding dominoes. Find the maximum value of $k(C)$ as a function of $n$.

Proposed by Holden Mui
83 replies
GoodMorning
Mar 23, 2023
ray66
Today at 7:30 AM
Titu Factoring Troll
GoodMorning   77
N Today at 7:19 AM by ray66
Source: 2023 USAJMO Problem 1
Find all triples of positive integers $(x,y,z)$ that satisfy the equation
$$2(x+y+z+2xyz)^2=(2xy+2yz+2zx+1)^2+2023.$$
77 replies
GoodMorning
Mar 23, 2023
ray66
Today at 7:19 AM
[Sipnayan SHS] Finals Round, Difficult
LilKirb   1
N Today at 6:59 AM by LilKirb
Let $f$ be a polynomial with nonnegative integer coefficients. If $f(1) = 11$ and $f(11) = 2311$, what is the remainder when $f(10)$ is divided by $1000?$
1 reply
LilKirb
Today at 6:49 AM
LilKirb
Today at 6:59 AM
Need help with combi problems
JARP091   4
N Today at 4:19 AM by JARP091
I want to create a problem set of some of the hardest combi problems that are yet to appear in any contest. Can anyone help me out? Also can anyone give me some tips to create combi problems.
4 replies
JARP091
Yesterday at 5:43 PM
JARP091
Today at 4:19 AM
Inequalities
sqing   1
N Today at 3:50 AM by sqing
Let $ a,b> 0 ,\frac{a}{2b+1}+\frac{b}{3}+\frac{1}{2a+1} \leq 1.$ Prove that
$$  a^2+b^2 -ab\leq 1$$$$ a^2+b^2 +ab \leq3$$Let $ a,b,c> 0 , \frac{a}{2b+1}+\frac{b}{2c+1}+\frac{c}{2a+1} \leq 1.$ Prove that
$$    a +b +c +abc \leq 4$$
1 reply
sqing
Today at 3:11 AM
sqing
Today at 3:50 AM
Inequalities
sqing   19
N Today at 2:50 AM by sqing
Let $ a,b>0   $ . Prove that
$$ \frac{a}{a^2+a +2b+1}+ \frac{b}{b^2+2a +b+1}  \leq  \frac{2}{5} $$$$ \frac{a}{a^2+2a +b+1}+ \frac{b}{b^2+a +2b+1}  \leq  \frac{2}{5} $$
19 replies
sqing
May 13, 2025
sqing
Today at 2:50 AM
inequality
luckvoltia.112   9
N Today at 2:35 AM by Shan3t
Given that \( a, b, c, d \) are nonzero real numbers, find the minimum value of the expression
\[
P = \left| \frac{b + c + d}{a} \right| + \left| \frac{c + d + a}{b} \right| + \left| \frac{d + a + b}{c} \right| + \left| \frac{a + b + c}{d} \right|.
\]
9 replies
luckvoltia.112
Today at 12:45 AM
Shan3t
Today at 2:35 AM
prove that
luckvoltia.112   0
Today at 12:52 AM
Let \( a, b, c \) be non-negative real numbers such that \( a + b + c > 0 \) and
\[
\frac{25a + 36b + 49c}{5a + 6b + 7c} + \frac{25b + 36c + 49a}{5b + 6c + 7a} + \frac{25c + 36a + 49b}{5c + 6a + 7b} = 18.
\]Prove that exactly two of the numbers \( a, b, c \) are equal to 0.
0 replies
luckvoltia.112
Today at 12:52 AM
0 replies
Geometry Trigonometry Olympiads
Foxellar   0
Yesterday at 11:07 PM
Let \( \triangle ABC \) be a triangle such that \( \angle ABC = 120^\circ \). Points \( X, Y, Z \) lie on segments \( BC, CA, AB \), respectively, such that lines \( AX, BY, \) and \( CZ \) are the angle bisectors of triangle \( ABC \). Find the measure of angle \( \angle XYZ \).
0 replies
Foxellar
Yesterday at 11:07 PM
0 replies
Proof of ramsey number
smadadi1000   1
N Yesterday at 10:35 PM by smadadi1000
How do you prove that r(n,2)=n using the pigeonhole principle?
1 reply
smadadi1000
Yesterday at 10:31 PM
smadadi1000
Yesterday at 10:35 PM
Minimize
lgx57   1
N Yesterday at 5:25 PM by Math-lover1
Minimize $\sqrt{\cos^2 x+(2-\sin x)^2}+\dfrac{1}{2}\sqrt{(\sqrt 3-\cos x)^2+(\sin x+1)^2}$
1 reply
lgx57
Yesterday at 1:29 PM
Math-lover1
Yesterday at 5:25 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
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
1011 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
1047 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
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megahertz13
3190 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
747 posts
#16
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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
1009 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
3752 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
3002 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
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#22
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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
91 posts
#23
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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
4243 posts
#24
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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
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$17_b$ 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
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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
3752 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
3059 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
1928 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
1319 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
1319 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
11379 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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