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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.

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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
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
Expressing polynomial as product of two polynomials
Sadigly   4
N 13 minutes ago by TBazar
Source: Azerbaijan Senior NMO 2021
Define $P(x)=((x-a_1)(x-a_2)...(x-a_n))^2 +1$, where $a_1,a_2...,a_n\in\mathbb{Z}$ and $n\in\mathbb{N^+}$. Prove that $P(x)$ couldn't be expressed as product of two non-constant polynomials with integer coefficients.
4 replies
Sadigly
Yesterday at 9:10 PM
TBazar
13 minutes ago
Interesting inequalities
sqing   8
N 17 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 0 , (a+k )(b+c)=k+1.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2k-3+2\sqrt{k+1}}{3k-1}$$Where $ k\geq \frac{2}{3}.$
Let $ a,b,c\geq 0 , (a+1)(b+c)=2.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 2\sqrt{2}-1$$Let $ a,b,c\geq 0 , (a+3)(b+c)=4.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{7}{4}$$Let $ a,b,c\geq 0 , (3a+2)(b+c)= 5.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2(2\sqrt{15}-5)}{3}$$
8 replies
1 viewing
sqing
May 10, 2025
sqing
17 minutes ago
Small Combinatorics Problem by Macharstein
macharstein   0
22 minutes ago
Alice and Bob are playing a two-turn game on an infinite grid. Alice goes first. In every cell, she places an arrow pointing in one of the four directions: up, down, left, or right. Then goes Bob. He puts $K$ robots, each in a different cell.
From this configuration the robots start moving: every second, each robot moves one step in the direction of the arrow in its current cell. If two robots land on the same cell at the same time, they merge into one.
Bob wins if the robots end up in the same row or same column infinitely many times.
Find all values of $K$ for which Alice can always prevent Bob from winning, no matter where he puts the robots.
0 replies
macharstein
22 minutes ago
0 replies
Shortest number theory you might've seen in your life
AlperenINAN   7
N 36 minutes ago by Assassino9931
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)+1$ is a perfect square, then $pq + 1$ is also a perfect square.
7 replies
AlperenINAN
Yesterday at 7:51 PM
Assassino9931
36 minutes ago
Help me this problem. Thank you
illybest   2
N 38 minutes ago by GreekIdiot
Find f: R->R such that
f( xy + f(z) ) = (( xf(y) + yf(x) )/2) + z
2 replies
illybest
3 hours ago
GreekIdiot
38 minutes ago
It just wants you to factorize 47 factorial
Sadigly   1
N 40 minutes ago by pooh123
Source: Azerbaijan Senior NMO 2021
At least how many numbers must be deleted from the product $1 \times 2 \times \dots \times 46 \times 47$ in order to make it a perfect square?
1 reply
Sadigly
Yesterday at 9:00 PM
pooh123
40 minutes ago
Inequality
Sadigly   5
N an hour ago by User141208
Source: Azerbaijan Senior NMO 2019
Prove that for any $a;b;c\in\mathbb{R^+}$, we have $$(a+b)^2+(a+b+4c)^2\geq \frac{100abc}{a+b+c}$$When does the equality hold?
5 replies
Sadigly
Yesterday at 8:47 PM
User141208
an hour ago
A strong inequality problem
hn111009   4
N an hour ago by hn111009
Source: Somewhere
Let $a,b,c$ be the positive number satisfied $a^2+b^2+c^2=3.$ Find the minimum of $$P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{3abc}{2(ab+bc+ca)}.$$
4 replies
hn111009
Yesterday at 2:02 AM
hn111009
an hour ago
Interesting inequalities
sqing   1
N an hour ago by pooh123
Source: Own
Let $ a,b>0  $ . Prove that
$$ \frac{a^2+b^2}{ab+1}+ \frac{4}{ (\sqrt{a}+\sqrt{b})^2} \geq 2$$$$ \frac{a^2+b^2}{ab+1}+ \frac{3}{a+\sqrt{ab}+b} \geq 2$$$$  \frac{a^3+b^3}{ab+1}+ \frac{4}{(a+b)^2}  \geq 2$$$$  \frac{a^3+b^3}{ab+1}+ \frac{3}{a^2+ab+b^2}  \geq 2$$$$\frac{a^2+b^2}{ab+2}+ \frac{1}{2\sqrt{ab}}  \geq \frac{2+3\sqrt{2}-2\sqrt{2(\sqrt{2}-1)}}{4} $$
1 reply
sqing
Today at 4:34 AM
pooh123
an hour ago
Diophantine involving cube
Sadigly   2
N an hour ago by Adywastaken
Source: Azerbaijan Senior NMO 2020
$a;b;c;d\in\mathbb{Z^+}$. Solve the equation: $$2^{a!}+2^{b!}+2^{c!}=d^3$$
2 replies
Sadigly
Yesterday at 10:13 PM
Adywastaken
an hour ago
Nice R+ FE
math_comb01   4
N an hour ago by mkultra42
Source: XOOK 2025 P3
Find all functions $f : \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ so that for any $x, y \in \mathbb{R}_{>0}$, we have \[ f(xf(x^2)+yf(x))=xf(y+xf(x)). \]
Proposed by Anmol Tiwari and MathLuis
4 replies
math_comb01
Feb 9, 2025
mkultra42
an hour ago
2018 JBMO TST- Macedonia, problem 4
Lukaluce   4
N an hour ago by Erto2011_
Source: 2018 JBMO TST- Macedonia
Determine all pairs $(p, q)$, $p, q \in \mathbb {N}$, such that

$(p + 1)^{p - 1} + (p - 1)^{p + 1} = q^q$.
4 replies
Lukaluce
May 28, 2019
Erto2011_
an hour ago
solve in positive integers: 3 \cdot 2^x +4 =n^2
parmenides51   4
N 2 hours ago by AylyGayypow009
Source: Greece JBMO TST 2019 p2
Find all pairs of positive integers $(x,n) $ that are solutions of the equation $3 \cdot 2^x +4 =n^2$.
4 replies
parmenides51
Apr 29, 2019
AylyGayypow009
2 hours ago
ISI UGB 2025 P7
SomeonecoolLovesMaths   10
N 2 hours ago by Mathworld314
Source: ISI UGB 2025 P7
Consider a ball that moves inside an acute-angled triangle along a straight line, unit it hits the boundary, which is when it changes direction according to the mirror law, just like a ray of light (angle of incidence = angle of reflection). Prove that there exists a triangular periodic path for the ball, as pictured below.

IMAGE
10 replies
SomeonecoolLovesMaths
Yesterday at 11:28 AM
Mathworld314
2 hours ago
All prime factors under 8
qwedsazxc   23
N Apr 22, 2025 by Giant_PT
Source: 2023 KMO Final Round Day 2 Problem 4
Find all positive integers $n$ satisfying the following.
$$2^n-1 \text{ doesn't have a prime factor larger than } 7$$
23 replies
qwedsazxc
Mar 26, 2023
Giant_PT
Apr 22, 2025
All prime factors under 8
G H J
G H BBookmark kLocked kLocked NReply
Source: 2023 KMO Final Round Day 2 Problem 4
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qwedsazxc
167 posts
#1 • 1 Y
Y by DEKT
Find all positive integers $n$ satisfying the following.
$$2^n-1 \text{ doesn't have a prime factor larger than } 7$$
This post has been edited 1 time. Last edited by qwedsazxc, Mar 26, 2023, 6:56 AM
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seojun8978
73 posts
#2
Y by
I took the test and I solved this problem. It needs some long calculations but they are not that hard. The answer seems to be 1,2,3,4,6
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Tintarn
9042 posts
#3 • 4 Y
Y by seojun8978, rightways, ljwn357, Assassino9931
Solution
This post has been edited 1 time. Last edited by Tintarn, Mar 26, 2023, 7:00 AM
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seojun8978
73 posts
#4 • 1 Y
Y by Tintarn
Tintarn wrote:
Solution
Nice Solution!! It's like 20 times shorter than my solution.
But.. Do you mean except?(expect)
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qwedsazxc
167 posts
#5
Y by
This was the easiest one on the test, probably everyone solved this.
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seojun8978
73 posts
#6 • 1 Y
Y by GuvercinciHoca
qwedsazxc wrote:
This was the easiest one on the test, probably everyone solved this.

Yes it's true. This would have been easier than geometry if there was one.
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Seungjun_Lee
526 posts
#7
Y by
seojun8978 wrote:
qwedsazxc wrote:
This was the easiest one on the test, probably everyone solved this.

Yes it's true. This would have been easier than geometry if there was one.

I think the geometry problem was as easy as this and maybe a little easier(on day 1)
This post has been edited 2 times. Last edited by Seungjun_Lee, Mar 26, 2023, 1:51 PM
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pokmui9909
185 posts
#8
Y by
I'm very curious about the prize cut. I heard there are a lot of participants who solved $4, 5$ problems.
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Seungjun_Lee
526 posts
#9
Y by
pokmui9909 wrote:
I'm very curious about the prize cut. I heard there are a lot of participants who solved $4, 5$ problems.

I did not take the test but it seems that 2 problems are given
Maybe really perfect 2problems or just 3problems will be honorable mention award
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IAmTheHazard
5001 posts
#10 • 2 Y
Y by megarnie, centslordm
Trivial by Zsigmondy
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Tintarn
9042 posts
#11
Y by
IAmTheHazard wrote:
Trivial by Zsigmondy
I agree that the problem is trivial and my first thought when I saw the problem was also Zsigmondy, but now I am quite certain that there is no detailed solution with Zsigmondy that is shorter than my elementary one in #3. (After all, there is no way around checking the small cases.)
This post has been edited 1 time. Last edited by Tintarn, Mar 26, 2023, 5:14 PM
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rightways
868 posts
#12
Y by
$Lemma:$

If $n>1$, then the greatest prime divisor of $2^n-1$ is greater than the greatest prime divisor of $n$.

Proof:

Let $p|n$ is greatest prime divisor of $n$, then for some prime $q$:
$q|2^p-1|2^n-1$, then $q>p$ because $p=ord_2(q)$. But $q$ is also a divisor of $2^n-1$, so it is not greater than gpd of $2^n-1$.

Now, in problem, by lemma we have that $7$ is greater than any prime divisor of $n$ , so we can write $n=2^a3^b5^c$ and we can easily finish problem by proving this $c<1$ and $b<2$ and $a<3$
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rightways
868 posts
#13
Y by
Can you post KJMO day 2 problems?
qwedsazxc wrote:
Find all positive integers $n$ satisfying the following.
$$2^n-1 \text{ doesn't have a prime factor larger than } 7$$
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qwedsazxc
167 posts
#14
Y by
rightways wrote:
Can you post KJMO day 2 problems?
qwedsazxc wrote:
Find all positive integers $n$ satisfying the following.
$$2^n-1 \text{ doesn't have a prime factor larger than } 7$$

There isn't a KJMO day 2. Final KMO contestants are picked from the people who excelled KMO 2nd round or KJMO 2nd round. I had the chance to take the Final KMO because I got a silver award from the KJMO I took on November 2022.
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hectorleo123
344 posts
#15
Y by
qwedsazxc wrote:
Find all positive integers $n$ satisfying the following.
$$2^n-1 \text{ doesn't have a prime factor larger than } 7$$
$2^2-1=2$
$2^3-1=7$
$2^4-1=3\times 5$
$gcd(2,1)=1$
By Zsigmondy's Theorem:
$\Rightarrow \exists$ prime $p \neq 2,3,5,7 / p|2^n-1, \forall n>4$
$\Rightarrow n\le 4$
But there is an exception $n=6$
$\Rightarrow n=1,2,3,4$ and $6$ are the only values that satisfy the condition $_\blacksquare$
This post has been edited 1 time. Last edited by hectorleo123, May 19, 2023, 1:48 AM
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Aiden-1089
285 posts
#16
Y by
It is easy to check that $n=1,2,3,4,6$ works. We claim that these are the only solutions.
Clearly $2 \nmid 2^n-1$. By Zsigmondy's theorem, any other $n$ would lead to $2^n-1$ having a prime divisor $p$ where $p \neq 3$ (because $3 \mid 2^2-1$), $5$ (because $5 \mid 2^4-1$), $7$ (because $7 \mid 2^3-1$). Since $p>7$, $n$ does not satisfy the condition.
This post has been edited 1 time. Last edited by Aiden-1089, Mar 26, 2024, 4:32 AM
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cursed_tangent1434
632 posts
#17 • 1 Y
Y by GeoKing
Me when this actually appeared on our National Olympiad. We claim that the only answers are $2,3,4$ and $6$. It is easy to see that these solutions indeed work. Now, we show that there are no other solutions.

We use the well known lemma that for all $d\mid n$ for positive integers $n$,
\[a^d - b^d \mid a^n - b^n\]for all positive integers $a,b$ extensively through out this solution. We first constrict the divisors of $n$.

Claim : There exists no prime divisor $p >3$ of $n$.
Proof : Say there exists such a prime factor $p$ of $n$. Then,
\[2^p -1 \mid 2^n-1\]and since $2^n-1$ has no prime factor larger than 7, this implies that $2^p-1$ also satisfies the same property. But, it is easy to see that $2 \nmid 2^n-1$ for any $n$, $3\mid 2^n-1$ only for even $n$, $5\mid 2^n-1$ only for $4\mid n$ and $7\mid 2^n-1$ only for $3\mid n$. Since $p>3$, this means that none of these prime factors can divide $2^p-1$ implying that it has a prime factor larger than 7, which is a clear contradiction. This proves the claim.

Now, we also note that,
\[2^8-1 = 255=3 \times 5 \times 17\]so, $8 \nmid n$ (since then $17 \mid 2^n-1$), and also
\[2^9 -1 = 511 = 7 \times 73\]so $9 \nmid n$ as well. This implies that $12 \mid n$ so $n \in \{1,2,3,4,6,12\}$. Now, we can simply check all these possibilities and see that they all work except for $1$ (which we exclude for conventional reasons) and $12$ (which we exclude since $2^{12}-1= 3^2 \times 5 \times 7 \times 13$ so $13$ is a prime factor), which implies that the solution set is indeed as claimed.
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kub-inst
31 posts
#18
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If $n$ satisfies the limit, then we can let $2^n-1=3^a\cdot 5^b\cdot 7^c$
Since$$3||(2^2-1),$$$$5||(2^4-1),$$$$7||(2^3-1)$$Then through LTE we can know that :
If $2|n,a=v_3(2^n-1)=v_3(\frac n2)+1\leq \log_3\frac n2+1.$
If $4|n,b=v_5(2^n-1)=v_5(\frac n4)+1\leq \log_5\frac n4+1.$
If $3|n,c=v_7(2^n-1)=v_7(\frac n3)+1\leq \log_7\frac n3+1.$
Hence, $$2^n-1=3^a\cdot 5^b\cdot 7^c\leq 3^{\log_3\frac n2+1}\cdot 5^{\log_5\frac n4+1}\cdot 7^{\log_7\frac n3+1}=\frac{85n^3}{24}$$(If $a, b$ or $c=0$, the inequality above is still ture evidently.)
Since $n\in \mathbb{N}^+,$ the inequality above requires $n\leq12$.
We can verify that the original statement is TRUE only when $n=1,2,3,4,6$ .$\square$
This post has been edited 1 time. Last edited by kub-inst, Jun 5, 2024, 6:03 AM
Reason: mistyped
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Scilyse
387 posts
#20
Y by
Zsigmondeez nuts
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Eka01
204 posts
#21
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By Zsigmondy, every $n$ gives a new prime factor apart from a small few (like greater than 10 or smth, dont have the energy to recall the edge cases). Then just check the few cases by hand.
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alexanderhamilton124
397 posts
#22
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7 | 2^3 - 1 ==> we done by zsigmondy
This post has been edited 1 time. Last edited by alexanderhamilton124, Jan 9, 2025, 3:34 AM
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RedFireTruck
4223 posts
#24
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n=1 gives 1
n=2 gives 3
n=3 gives 7
n=4 gives 15=3*5
n=6 gives 63=3*3*7

by zsigmondy theorem, every other 2^n-1 must have a prime factor other than 3,5,7 so this is all
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Ihatecombin
60 posts
#25
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L.T.E also works here, even with brain dead level bounding one only needs to check until $n=14$.
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Giant_PT
31 posts
#26
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Almost trivial by Zsigmondy :(
We can see that the smallest $n$ for which $3|2^{n}-1$, $5|2^{n}-1$, and $7|2^{n}-1$ are $2$, $4$ and $3$ respectively. Also, we have the exceptional case when $n=6$, so by just checking small values of $n$, we see that $n=1,2,3,4,6$ are the only solutions.
This post has been edited 2 times. Last edited by Giant_PT, Apr 22, 2025, 3:41 PM
Reason: Typos
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