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k a My Retirement & New Leadership at AoPS
rrusczyk   1571
N Mar 26, 2025 by SmartGroot
I write today to announce my retirement as CEO from Art of Problem Solving. When I founded AoPS 22 years ago, I never imagined that we would reach so many students and families, or that we would find so many channels through which we discover, inspire, and train the great problem solvers of the next generation. I am very proud of all we have accomplished and I’m thankful for the many supporters who provided inspiration and encouragement along the way. I'm particularly grateful to all of the wonderful members of the AoPS Community!

I’m delighted to introduce our new leaders - Ben Kornell and Andrew Sutherland. Ben has extensive experience in education and edtech prior to joining AoPS as my successor as CEO, including starting like I did as a classroom teacher. He has a deep understanding of the value of our work because he’s an AoPS parent! Meanwhile, Andrew and I have common roots as founders of education companies; he launched Quizlet at age 15! His journey from founder to MIT to technology and product leader as our Chief Product Officer traces a pathway many of our students will follow in the years to come.

Thank you again for your support for Art of Problem Solving and we look forward to working with millions more wonderful problem solvers in the years to come.

And special thanks to all of the amazing AoPS team members who have helped build AoPS. We’ve come a long way from here:IMAGE
1571 replies
rrusczyk
Mar 24, 2025
SmartGroot
Mar 26, 2025
k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 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
Inequality
JK1603JK   0
9 minutes ago
Let $a,b,c\ge 0: a+b+c=3$ then prove \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2}\cdot\frac{abc}{ab+bc+ca}\ge \frac{5}{3}.$$
0 replies
JK1603JK
9 minutes ago
0 replies
inequalities hard
Cobedangiu   1
N 37 minutes ago by Cobedangiu
problem
1 reply
Cobedangiu
Monday at 11:45 AM
Cobedangiu
37 minutes ago
Something nice
KhuongTrang   25
N 39 minutes ago by KhuongTrang
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
25 replies
KhuongTrang
Nov 1, 2023
KhuongTrang
39 minutes ago
Inequality
JK1603JK   1
N 43 minutes ago by lbh_qys
Prove that 9ab\left(a-b+c\right)+9bc\left(b-c+a\right)+9ca\left(c-a+b\right)\ge \left(a+b+c\right)^{3},\ \ a\ge 0\ge b\ge c: a+b+c\le 0.
1 reply
JK1603JK
an hour ago
lbh_qys
43 minutes ago
inequalities
Cobedangiu   7
N an hour ago by arqady
problem
7 replies
Cobedangiu
Mar 31, 2025
arqady
an hour ago
Symmedian tangent
hsiangshen   13
N an hour ago by imzzzzzz
Source: My friend
Let $O,K$ be the circumcenter, symmedian point of $\triangle ABC$. Show that the tangent of $(AOK)$ at $A$,the tangent of $(BOK)$ at $B$, the tangent of $(COK)$ at $C$ are concurrent.
13 replies
hsiangshen
Jan 6, 2021
imzzzzzz
an hour ago
Solve the equetion
yt12   4
N an hour ago by lgx57
Solve the equetion:$\sin 2x+\tan x=2$
4 replies
yt12
Mar 31, 2025
lgx57
an hour ago
Inequalities
sqing   2
N 2 hours ago by sqing
Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=11.$ Prove that
$$a+ab+abc\leq\frac{49}{6}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=10.$ Prove that
$$a+ab+abc\leq\frac{169}{24}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=14.$ Prove that
$$a+ab+abc\leq\frac{63+5\sqrt 5}{6}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=32.$ Prove that
$$a+ab+abc\leq48+\frac{64\sqrt{2}}{3}$$
2 replies
sqing
Yesterday at 2:59 PM
sqing
2 hours ago
2023 factors and perfect cube
proxima1681   5
N 2 hours ago by mqoi_KOLA
Source: Indian Statistical Institute (ISI) UGB 2023 P4
Let $n_1, n_2, \cdots , n_{51}$ be distinct natural numbers each of which has exactly $2023$ positive integer factors. For instance, $2^{2022}$ has exactly $2023$ positive integer factors $1,2, 2^{2}, 2^{3}, \cdots 2^{2021}, 2^{2022}$. Assume that no prime larger than $11$ divides any of the $n_{i}$'s. Show that there must be some perfect cube among the $n_{i}$'s.
5 replies
proxima1681
May 14, 2023
mqoi_KOLA
2 hours ago
Integer Symmetric Polynomials
proxima1681   4
N 2 hours ago by mqoi_KOLA
Source: Indian Statistical Institute (ISI) UGB 2023 P7
(a) Let $n \geq 1$ be an integer. Prove that $X^n+Y^n+Z^n$ can be written as a polynomial with integer coefficients in the variables $\alpha=X+Y+Z$, $\beta= XY+YZ+ZX$ and $\gamma = XYZ$.
(b) Let $G_n=x^n \sin(nA)+y^n \sin(nB)+z^n \sin(nC)$, where $x,y,z, A,B,C$ are real numbers such that $A+B+C$ is an integral multiple of $\pi$. Using (a) or otherwise show that if $G_1=G_2=0$, then $G_n=0$ for all positive integers $n$.
4 replies
proxima1681
May 14, 2023
mqoi_KOLA
2 hours ago
IMO 2008, Question 1
orl   153
N 2 hours ago by QueenArwen
Source: IMO Shortlist 2008, G1
Let $ H$ be the orthocenter of an acute-angled triangle $ ABC$. The circle $ \Gamma_{A}$ centered at the midpoint of $ BC$ and passing through $ H$ intersects the sideline $ BC$ at points $ A_{1}$ and $ A_{2}$. Similarly, define the points $ B_{1}$, $ B_{2}$, $ C_{1}$ and $ C_{2}$.

Prove that the six points $ A_{1}$, $ A_{2}$, $ B_{1}$, $ B_{2}$, $ C_{1}$ and $ C_{2}$ are concyclic.

Author: Andrey Gavrilyuk, Russia
153 replies
orl
Jul 16, 2008
QueenArwen
2 hours ago
P(x) doesn't have 2024 distinct real roots
gatnghiep   1
N 2 hours ago by kiyoras_2001
Given a polynomial $P(x)=x^{2024}+a_{2023}x^{2023}+...+a_1x+1$ with real coefficient. It is known that $|a_{1012}|<2$ and $a_k = a_{2024-k}, \forall k = 1,2,...,2012$. Prove that $P(x)$ can't have $2024$ distinct real roots.
1 reply
gatnghiep
Oct 2, 2024
kiyoras_2001
2 hours ago
A lies on the radical axis of BQX and CPX
a_507_bc   35
N 3 hours ago by jordiejoh
Source: APMO 2024 P1
Let $ABC$ be an acute triangle. Let $D$ be a point on side $AB$ and $E$ be a point on side $AC$ such that lines $BC$ and $DE$ are parallel. Let $X$ be an interior point of $BCED$. Suppose rays $DX$ and $EX$ meet side $BC$ at points $P$ and $Q$, respectively, such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $BQX$ and $CPX$ intersect at a point $Y \neq X$. Prove that the points $A, X$, and $Y$ are collinear.
35 replies
a_507_bc
Jul 29, 2024
jordiejoh
3 hours ago
sum(ab/4a^2+b^2) <= 3/5
truongphatt2668   3
N 3 hours ago by Nguyenhuyen_AG
Source: I remember I read it somewhere
Let $a,b,c>0$. Prove that:
$$\dfrac{ab}{a^2+4b^2} + \dfrac{bc}{b^2+4c^2} + \dfrac{ca}{c^2+4a^2} \le \dfrac{3}{5}$$
3 replies
truongphatt2668
Monday at 1:23 PM
Nguyenhuyen_AG
3 hours ago
AIME Problem
paladin8   19
N Jul 15, 2022 by fidgetboss_4000
Let $A$, $B$, $C$, and $D$ be the vertices of a regular tetrahedron, each of whose edges measures 1 meter. A bug, starting from vertex $A$, observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = n/729$ be the probability that the bug is at vertex $A$ when it has crawled exactly 7 meters. Find the value of $n$.
19 replies
paladin8
Nov 25, 2003
fidgetboss_4000
Jul 15, 2022
AIME Problem
G H J
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paladin8
3237 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $A$, $B$, $C$, and $D$ be the vertices of a regular tetrahedron, each of whose edges measures 1 meter. A bug, starting from vertex $A$, observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = n/729$ be the probability that the bug is at vertex $A$ when it has crawled exactly 7 meters. Find the value of $n$.
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ComplexZeta
2853 posts
#2 • 3 Y
Y by Adventure10, Mango247, and 1 other user
With a bit of intuition, you can actually solve this problem with practically no math. After the bug has been going for a while, it will have roughly 1/4 chance of being at any of the four vertices. Since the probability is n/729, n is close to 729/4, or 182.25. 182 is your answer.

Now, that isn't very rigorous, and I checked that it was indeed correct by doing it properly. Nonetheless, if you don't know how to do a problem like this on the AIME or any other contest, this argument will almost always give you the correct answer. (I have yet to see a problem like this where this method doesn't work.)

I think I'll let someone else solve this problem properly before posting my solution.
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JBL
16123 posts
#3 • 2 Y
Y by Adventure10, Mango247
Wow, that's really cool. I'm sure it's possible to formulate a threshold for the number of moves on a given graph necessary to make that probability as close as it needs to be. A very interesting project, actually. Did you think that method up yourself, or did someone show it to you?
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TripleM
1587 posts
#4 • 2 Y
Y by Adventure10, Mango247
Well seeing that I've done work on markov chains I could get an easy solution using that logic



But as for a simple solution.. well - I'm doing this as I go along so this message may not even end up being posted, but if you're reading this it must have been.. anyway..



Click to reveal hidden text
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ComplexZeta
2853 posts
#5 • 2 Y
Y by Adventure10, Mango247
I came up with it myself after seeing way too many problems with answers round(2^n/3)/2^n.
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MysticTerminator
3697 posts
#6 • 2 Y
Y by Adventure10, Mango247
or, from MMM's recursion Click to reveal hidden text we derive the closed form Click to reveal hidden text.
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fairyofwind
156 posts
#7 • 2 Y
Y by Adventure10, Mango247
set it up as a communication network take powers of matrix to get number of paths through however many relays. divide.
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TripleM
1587 posts
#8 • 2 Y
Y by Adventure10, Mango247
Yeah, thats what I was saying about markov chains, same thing. But for this type of problem it is a lot easier to do it some other way.
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joml88
6343 posts
#9 • 2 Y
Y by Adventure10, Mango247
Can someone post a solution such that the post contains just the solution and no extraneous remarks?

Thanks

P.S. I'll delete this post after someone posts a solution.
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mna851
549 posts
#10 • 2 Y
Y by Adventure10, Mango247
Let $\ P_n$ be the probability that the bug is on vertex $\ A$ after $\ n$ moves. Clearly, this means $\ P_1 = 0$ and $\ P_2= \frac{1}{3}$ We can also define the recursion $\ P_n = 1/3(1- P_{n-1})$ , because on the move prior to move $\ n$ the bug must be at a point other than $\ A$. Now starting from $\ P_2 = 1/3$ we can repeat this recursion up to $\ P_7$ doing so we get $\ P_3 = \frac{2}{9}$ $\ P_4 = \frac{7}{27}$ $\ P_5 = \frac{20}{81}$ $\ P_6 = \frac{61}{243}$ and finally $\ P_7 = \frac{182}{729}$ So $\ n=182$
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Euler2718
2 posts
#11 • 2 Y
Y by Adventure10, Mango247
I did this with graph theory:
The tetrahedron is basically the graph K4, and has an adjacency matrix with 0's along its main diagonal and 1's everywhere else. Raise this matrix to the 7th power to find the number of paths of length 7 from any given vertex to another. This matrix has 546's along its main diagonal and 547's everywhere else. The number of length 7 A-A paths is then 546. Since we must start at A, our sample space lies in the first row of the matrix, which tells us that there are 546+3*547 paths of length 7 beginning at A. Therefore, the probability that a path of length 7 starting at A also ends at A is 546/(546+3*547) = 182/729, so n=182.
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mathwizarddude
1976 posts
#12 • 2 Y
Y by Adventure10, Mango247
Could someone please explain how the solution with graph theory (and matrices) work? Thanks!
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grn_trtle
1281 posts
#13 • 2 Y
Y by Adventure10, Mango247
I learned today how Markov chains worked in my school's math club, which immediately reminded me of this problem, so I'm going to take a crack at it. (This might answer your question, mathwizarddude, about a matrix solution to this problem).

Solution with Markov chains
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t0rajir0u
12167 posts
#14 • 2 Y
Y by Adventure10, Mango247
grn_trtle wrote:
Doing it by hand on the actual AIME is a piece of... steak—chewy, but delicious ;).
Proposition: The eigenvalues of the matrix you wrote down are $ 1, - \frac {1}{3}, - \frac {1}{3}, - \frac {1}{3}$.

Lemma: Let $ P$ be a stochastic matrix. The probability that after $ n$ steps you return to the same state as the state in which you started is the sum of the $ n^{th}$ powers of the eigenvalues of $ P$.

So...
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grn_trtle
1281 posts
#15 • 2 Y
Y by Adventure10, Mango247
Very nice :)

Are there similar ways to find certain elements of the matrix after $ n$ steps? (If you wanted find the probability of going from one state to another state in $ n$ steps, without working out matrix exponentiation)
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t0rajir0u
12167 posts
#16 • 2 Y
Y by Adventure10, Mango247
In general you'd either need to work directly with the matrix or compute eigenvectors. Closed walks are unique because they depend on the eigenvalues only. There is a theorem which tells you how to do this without computing eigenvectors, but it is somewhat indirect.

Theorem (4.7.2 in Stanley, Enumerative Combinatorics): For an adjacency matrix $ A$, let $ a_{ij}(n)$ denote the number of walks of length $ n$ from vertex $ i$ to vertex $ j$ and let $ A[ij]$ denote the matrix obtained from $ A$ by deleting the $ i^{th}$ column and $ j^{th}$ row. Then

$ \sum_{n \ge 0} a_{ij}(n) \lambda^n = \frac {( - 1)^{i + j} \det(I - \lambda A[ij]) }{\det(I - \lambda A) }$.
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OlympusHero
17019 posts
#17 • 1 Y
Y by WicMic
Note that from each vertex you can go to any of the other three. Now we go step by step.

First meter: 0 chance of going to A, 1 chance of going to B, C, D
Second meter: 1/3 chance of going to A, 2/3 chance of going to B, C, D
Third meter: 2/9 chance of going to A, 7/9 chance of going to B, C, D
Fourth meter: 7/27 chance of going to A, 20/27 chance of going to B, C, D
Fifth meter: 20/81 chance of going to A, 61/81 chance of going to B, C, D
Sixth meter: 61/243 chance of going to A, 182/243 chance of going to B, C, D
Seventh meter: $\boxed{\frac{182}{729}}$ chance of going to A.

Wow.
This post has been edited 2 times. Last edited by OlympusHero, Jan 22, 2022, 7:01 PM
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OlympusHero
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#18
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The answer to 2022 AMC 8 P25 is contained within this post.
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andlind
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#19
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They are basically the same problem with different words with this one going longer so it makes sense that they go through the same numbers.
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fidgetboss_4000
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#20
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Solution
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N Quick Reply
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a