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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.

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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
Inspired by m4thbl3nd3r
sqing   1
N 27 minutes ago by sqing
Source: Own
Let $  a, b,c>0,b+c>a$. Prove that$$\sqrt{\frac{a}{b+c-a}}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)}\geq 1$$$$\frac{a}{b+c-a}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)} \geq  \frac{4\sqrt 2}{3}-1$$
1 reply
sqing
42 minutes ago
sqing
27 minutes ago
2-var inequality
sqing   6
N an hour ago by sqing
Source: Own
Let $ a,b> 0 $ and $2a+2b+ab=5. $ Prove that
$$ \frac{a^2}{b^2}+\frac{1}{a^2}-a^2\geq  1$$$$ \frac{a^3}{b^3}+\frac{1}{a^3}-a^3\geq  1$$
6 replies
sqing
Yesterday at 1:19 PM
sqing
an hour ago
Inequality
Amin12   8
N an hour ago by A.H.H
Source:  Iran 3rd round-2017-Algebra final exam-P3
Let $a,b$ and $c$ be positive real numbers. Prove that
$$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $$
8 replies
Amin12
Sep 2, 2017
A.H.H
an hour ago
Not so beautiful
m4thbl3nd3r   0
an hour ago
Let $a, b,c>0$ such that $b+c>a$. Prove that $$2 \sqrt[4]{\frac{a}{b+c-a}}\ge 2 +\frac{2a^2-b^2-c^2}{(a+b)(a+c)}.$$
0 replies
m4thbl3nd3r
an hour ago
0 replies
Every subset of size k has sum at most N/2
orl   50
N 2 hours ago by de-Kirschbaum
Source: USAMO 2006, Problem 2, proposed by Dick Gibbs
For a given positive integer $k$ find, in terms of $k$, the minimum value of $N$ for which there is a set of $2k + 1$ distinct positive integers that has sum greater than $N$ but every subset of size $k$ has sum at most $\tfrac{N}{2}.$
50 replies
orl
Apr 20, 2006
de-Kirschbaum
2 hours ago
Inspired by a9opsow_
sqing   2
N 2 hours ago by sqing
Source: Own
Let $ a,b > 0  .$ Prove that
$$ \frac{(ka^2 - kab-b)^2 + (kb^2 - kab-a)^2 + (ab-ka-kb )^2}{ (ka+b)^2 + (kb+a)^2+(a - b)^2 }\geq  \frac {1}{(k+1)^2}$$Where $ k\geq 0.37088 .$
$$\frac{(a^2 - ab-b)^2 + (b^2 - ab-a)^2 + ( ab-a-b)^2}{a^2 +b^2+(a - b)^2 } \geq 1$$$$ \frac{(2a^2 - 2ab-b)^2 + (2b^2 - 2ab-a)^2 + (ab-2a-2b )^2}{ (2a+b)^2 + (2b+a)^2+(a - b)^2 }\geq  \frac 19$$
2 replies
sqing
3 hours ago
sqing
2 hours ago
Cute NT Problem
M11100111001Y1R   5
N 3 hours ago by compoly2010
Source: Iran TST 2025 Test 4 Problem 1
A number \( n \) is called lucky if it has at least two distinct prime divisors and can be written in the form:
\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]where \( p_1, \dots, p_k \) are distinct prime numbers that divide \( n \). (Note: it is possible that \( n \) has other prime divisors not among \( p_1, \dots, p_k \).) Prove that for every prime number \( p \), there exists a lucky number \( n \) such that \( p \mid n \).
5 replies
M11100111001Y1R
Tuesday at 7:20 AM
compoly2010
3 hours ago
3 var inequality
SunnyEvan   13
N 3 hours ago by Nguyenhuyen_AG
Let $ a,b,c \in R $ ,such that $ a^2+b^2+c^2=4(ab+bc+ca)$Prove that :$$ \frac{7-2\sqrt{14}}{48} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{7+2\sqrt{14}}{48} $$
13 replies
SunnyEvan
May 17, 2025
Nguyenhuyen_AG
3 hours ago
trigonometric inequality
MATH1945   13
N 4 hours ago by sqing
Source: ?
In triangle $ABC$, prove that $$sin^2(A)+sin^2(B)+sin^2(C) \leq \frac{9}{4}$$
13 replies
MATH1945
May 26, 2016
sqing
4 hours ago
Iran TST Starter
M11100111001Y1R   2
N 4 hours ago by sami1618
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[
a_n < \frac{1}{1404}
\]
2 replies
M11100111001Y1R
Tuesday at 7:36 AM
sami1618
4 hours ago
Twin Prime Diophantine
awesomeming327.   23
N 5 hours ago by HDavisWashu
Source: CMO 2025
Determine all positive integers $a$, $b$, $c$, $p$, where $p$ and $p+2$ are odd primes and
\[2^ap^b=(p+2)^c-1.\]
23 replies
awesomeming327.
Mar 7, 2025
HDavisWashu
5 hours ago
Troublesome median in a difficult inequality
JG666   2
N 5 hours ago by navid
Source: 2022 Spring NSMO Day 2 Problem 3
Determine the minimum value of $\lambda\in\mathbb{R}$, such that for any positive integer $n$ and non-negative reals $x_1, x_2, \cdots, x_n$, the following inequality always holds:
$$\sum_{i=1}^n(m_i-a_i)^2\leqslant \lambda\cdot\sum_{i=1}^nx_i^2,$$Here $m_i$ and $a_i$ denote the median and arithmetic mean of $x_1, x_2, \cdots, x_i$, respectively.

Duanyang ZHANG, High School Affiliated to Renmin University of China
2 replies
1 viewing
JG666
May 22, 2022
navid
5 hours ago
Concurrent lines, angle bisectors
legogubbe   0
5 hours ago
Source: ???
Hi AoPS!

Let $ABC$ be an isosceles triangle with $AB=AC$, and $M$ an arbitrary point on side $BC$. The internal angle bisector of $\angle MAB$ meets the circumcircle of $\triangle ABC$ again at $P \neq A$, and the internal angle bisector of $\angle CAM$ meets it again at $Q \neq A$. Show that lines $AM$, $BQ$ and $CP$ are concurrent.
0 replies
legogubbe
5 hours ago
0 replies
Fractional Inequality
sqing   33
N 5 hours ago by Learning11
Source: Chinese Girls Mathematical Olympiad 2012, Problem 1
Let $ a_1, a_2,\ldots, a_n$ be non-negative real numbers. Prove that
$\frac{1}{1+ a_1}+\frac{ a_1}{(1+ a_1)(1+ a_2)}+\frac{ a_1 a_2}{(1+ a_1)(1+ a_2)(1+ a_3)}+$ $\cdots+\frac{ a_1 a_2\cdots a_{n-1}}{(1+ a_1)(1+ a_2)\cdots (1+ a_n)} \le 1.$
33 replies
sqing
Aug 10, 2012
Learning11
5 hours ago
2024 IMO P1
EthanWYX2009   103
N Apr 27, 2025 by ashwinmeena
Source: 2024 IMO P1
Determine all real numbers $\alpha$ such that, for every positive integer $n,$ the integer
$$\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor$$is a multiple of $n.$ (Note that $\lfloor z\rfloor$ denotes the greatest integer less than or equal to $z.$ For example, $\lfloor -\pi\rfloor =-4$ and $\lfloor 2\rfloor= \lfloor 2.9\rfloor =2.$)

Proposed by Santiago Rodríguez, Colombia
103 replies
EthanWYX2009
Jul 16, 2024
ashwinmeena
Apr 27, 2025
2024 IMO P1
G H J
G H BBookmark kLocked kLocked NReply
Source: 2024 IMO P1
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almagest3001
6 posts
#99 • 1 Y
Y by cubres
The solution is all even $\alpha \in \mathbb{Z}$. Indeed that works: $n\mid \frac{n(n+1)}{2}\alpha$.
If $\alpha \in \mathbb{Z}$ is odd, pick $n$ even and $(n, \alpha) =1$, then $n \nmid \frac{n(n+1)}{2}\alpha$.
Suppose $\alpha \in \mathbb{R} \setminus \mathbb{Z}$. Then pick $n$ such that $n\{\alpha\} \ge 1$ but $(n-1)\{\alpha\}  < 1$. Because $n \ge 2$, $\lfloor \alpha \rfloor \ne 0$. Else, if $n = 2k$, $k \nmid k(2k+1)\lfloor \alpha \rfloor +1$, if $n = 2k+1$, $2k+1 \nmid (2k+1)(k+1)\lfloor \alpha \rfloor +1$ so this case also doesn't work out.
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Antonyliao
15 posts
#101 • 1 Y
Y by cubres
Very interesting solution...(Not sure if anyone sees this, but anyways,)

Claim: Solution is all even $\alpha \in \mathbb{Z}$.
Proof: assume that $k \in \mathbb{Z}$. Then we obviously have
$\lfloor {\alpha}\rfloor + \lfloor {2\alpha} \rfloor + . . . + \lfloor {k\alpha} \rfloor \equiv 0 \pmod{k}$
and
$\lfloor {\alpha}\rfloor + \lfloor {2\alpha} \rfloor + . . . + \lfloor {k\alpha} \rfloor \equiv -\lfloor {(k+1)\alpha}\rfloor \pmod{k+1}$ . Thus,
we can express $\lfloor {\alpha}\rfloor + \lfloor {2\alpha} \rfloor + . . . + \lfloor {k\alpha} \rfloor = kx$, where $x \in \mathbb{Z}$ . So,
$kx \equiv -x \equiv -\lfloor {(k+1)\alpha}\rfloor \pmod{k+1}$ . We can express $x = m(k+1) + \lfloor {(k+1)a} \rfloor$, where $m \in \mathbb{Z}$.

Assume $0 \le \alpha < 1$. Notice that $\lfloor {\alpha}\rfloor + \lfloor {2\alpha} \rfloor + . . . + \lfloor {k\alpha} \rfloor = kx < k^2 \Longrightarrow 0 \le x < k$ , so $m$ must be equal to 0. So $x = \lfloor {\alpha}\rfloor + \lfloor {2\alpha} \rfloor + . . . + \lfloor {k\alpha} \rfloor = \lfloor {(k+1)a} \rfloor$. by $ \lfloor {\alpha}\rfloor + \lfloor {2\alpha} \rfloor + . . . + \lfloor {(k+1)\alpha} \rfloor = \lfloor {(k+2)a} \rfloor$, we get that
$ \lfloor {k\alpha} \rfloor  = \lfloor {(k+1)\alpha} \rfloor $ for all k, which means $\alpha = 0$.

Now, consider solutions not in the interval $0 \le \alpha < 1$. Its easy to see that $\lfloor {\alpha}\rfloor + \lfloor {2\alpha} \rfloor + . . . + \lfloor {n\alpha} \rfloor + sn(n+1)/2  = \lfloor {\alpha+s}\rfloor + \lfloor {2\alpha+2s} \rfloor + . . . + \lfloor {n\alpha+ns} \rfloor$ for any $s \in \mathbb{Z}$. Thus, by knowing the values of $\lfloor {\alpha}\rfloor + \lfloor {2\alpha} \rfloor + . . . + \lfloor {k\alpha} \rfloor$ in $0 \le \alpha < 1$ , all values in $\mathbb {R} $ can be computed. In order for there to be more solutions other than $\alpha = 0$, $sn(n+1)/2$ should be divisible by n $\Longrightarrow s \equiv 0 \pmod{2}$. Therefore if $\alpha$ is a solution, $\alpha + 2$ is also one. Therefore, the claim is true. $\blacksquare$
This post has been edited 2 times. Last edited by Antonyliao, Sep 10, 2024, 1:29 PM
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Cali.Math
128 posts
#102 • 1 Y
Y by cubres
We uploaded our solution https://calimath.org/pdf/IMO2024-1.pdf on youtube https://youtu.be/ewvE5_kNw1I.
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kurage
2 posts
#103 • 3 Y
Y by GeorgeRP, segment, cubres
α obviously satisfies when α is an even number.
Assuming α is not an integer, let m = [α]. Then, there is a positive integer k such that m+1/(k+1) <= α < m + 1/k. The summation is n(n-1)m/2 for n < k + 1 and n(n-1)m/2 + 1 for n = k + 1. We notice that m is an even number since n divides the summation for n < k + 1. Therefore, the summation can be written as n(n-1)m' + 1 for n = k + 1 where m = 2m', which is relatively prime to n.
Finally, the answers {α} are set of even numbers.
This post has been edited 1 time. Last edited by kurage, Sep 18, 2024, 2:11 AM
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asqb
10 posts
#104 • 1 Y
Y by cubres
\documentclass{article}
\usepackage{amsmath}

\begin{document}

\textbf{1-Masala:} $\alpha$ haqiqiy sonning barcha qiymatlarini topingki, bunda $n$ musbat butun sonning har bir qiymatida
\[
\lfloor \alpha \rfloor + \lfloor 2\alpha \rfloor + \cdots + \lfloor n\alpha \rfloor
\]ifoda $n$ ga qoldiqsiz bo’lsin. (Izoh: $\lfloor z \rfloor$ orqali $z$ dan kichik yoki teng bo’lgan eng katta butun sonni belgilaymiz. Masalan, $\lfloor -\pi \rfloor = -4$ va $\lfloor 2.9 \rfloor = 2$.)

\end{document}
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Golden_Verse
5 posts
#105 • 1 Y
Y by cubres
Answer
Solution
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megahertz13
3194 posts
#106 • 2 Y
Y by kilobyte144, cubres
The answer is even integers.

There are two cases.

Case 1: $\alpha$ is an integer. We know that $n$ is a factor of $$\alpha+2\alpha+\dots+n\alpha=\alpha\frac{n(n-1)}{2},$$so $$\alpha(\frac{n-1}{2})$$must be an integer. Since $n$ can be an even number, $\alpha$ must be an even integer here.

Case 2: $\alpha$ is not an integer. We can translate $\alpha$ by multiples of $2$ until $-1<\alpha<1$.

Note that $$\lfloor \alpha+2k \rfloor + \lfloor 2(\alpha+2k) \rfloor + \dots + \lfloor n(\alpha+2k) \rfloor \equiv \lfloor \alpha \rfloor + \lfloor 2\alpha \rfloor + \dots + \lfloor n\alpha \rfloor\pmod n$$if $k$ is an integer.

Case 2.1: $0<\alpha<1$. Let $k$ be the smallest positive integer satisfying $k\alpha > 1$. Setting $n=k$, we have $$0+0+\dots+0+0+1$$is a multiple of $n$. This implies that $n=k=1$, so $\alpha>1$, a contradiction.

Case 2.2: $-1<\alpha<0$. Let $k$ be the smallest positive integer satisfying $k\alpha < -1$. Setting $n=k$, we have $$(-1)+(-1)+\dots+(-1)+(-1)+(-2)=(-1)(n-1)+(-2)=-n-1.$$Since this is a multiple of $n$, we know that $n = k = 1$, so $\alpha<-1$, a contradiction.
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lelouchvigeo
183 posts
#107 • 1 Y
Y by cubres
sketch
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onyqz
195 posts
#108 • 1 Y
Y by cubres
storage
solution
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eg4334
636 posts
#109
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The answer is only $\boxed{\text{even integers a}}$. These obviously work. Shift $a$ to the interval $[0, 2]$ because for anything larger we can take out a $n(n+1)$ from the sum of floors which dies mod $n$. In general the divisibility condition for $n$ is satisfied for $a$ some interval. Call this the good interval of $a$. If this is true then the condition obviously follows by taking a sufficiently large $n$. We wish to prove that the intersection of the first $n$ good intervals for any $n$ is the set $[0, \frac{1}{n}) \cup [2 - \frac{1}{n}, 2]$. We prove this using induction. If the statement is true for some $N$, we wish to prove $N+1$ because of trivial base case. Notice that the good interval contains $[0, \frac{1}{n+1})$ because the sum is zero. Now we need to prove that everything in $[\frac{1}{n+1}, \frac{1}{n})$ fails because the other side is basically the same reasoning. This is true because the sum here has the 2nd to last term in the sum equal to zero still but the last term is not enough to push it to a multiple of $n+1$.
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Maximilian113
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#110
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Observe that shifting $\alpha$ by $2$ yields a change of $n(n+1)$ so if $\alpha$ works then $\alpha \pm 2$ also does. Hence WLOG assume that $-1 \leq \alpha \leq 1.$ It is easy to see that $\alpha=\pm 1$ does not work by setting $n=2.$ If $\alpha=0,$ it clearly works.

Now assume that $\alpha$ is not an integer. If $\alpha > 0,$ let $m$ be the smallest positive integer such that $m\alpha \geq 1.$ Then setting $n=m$ yields $m | 1,$ impossible. Similarly $\alpha < 0$ does not work too.

Therefore, only $\alpha=0$ works so the answer is all even integers $\alpha.$
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HamstPan38825
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#111
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Let $\alpha = k + r$ for $k = \lfloor \alpha \rfloor$, such that for all positive integers $m$, $m\alpha = mk + mr$ has integer part $mk + \lfloor mr \rfloor$. We denote $S_n(\alpha)$ to denote the given sum truncated at $n$ terms with argument $\alpha$.

In particular, let $2n+1 \geq 3$ be an odd integer. Then $2n+1$ divides $S_{2n+1}(\alpha)$ if and only if it also divides $S_{2n+1}(r)$. Now, assume that $r \neq 0$, and we split into two cases:

First Case: Suppose that $r < \tfrac 12$. Then, let $2n+1$ be the smallest odd integer such that $(2n+1)r > 1$, so that $ir < 1$ for all $1 \leq i \leq 2n-1$. Thus $(2n+1)r < 2$, and it follows that $S_{2n+1}(r) \leq 2$ as only the $2n$ and $2n+1$-coefficient terms are nonzero. Thus $2n+1 \nmid S_{2n+1}(r)$, so such $r$ cannot satisfy the conditions.

Second Case: Suppose that $r > \tfrac 12$. Then, let $2n+1$ be the smallest odd integer such that $(2n+1)r < 2n$, so that $\lfloor ir \rfloor = i-1$ for all $1 \leq i \leq 2n-1$ as $r < 1$ too. It follows that \[S_{2n+1}(r) = 1 + 2 + \cdots + 2n-2+ 2n-1 + 2n-1 - \varepsilon = n(2n+1) - 1 - \varepsilon\]for some $\varepsilon \in \{0, 1\}$ depending on the value of $\lfloor 2n r \rfloor \in \{2n-2, 2n-1\}$. Since $2n+1 \geq 3$, this is also not a multiple of $2n+1$.

So $r = 0$, and $\alpha$ must be an integer. Clearly we see that $\alpha$ must be an even integer now.

Remark: I expected this problem to have something of a nice solution, but I guess not. It is what it is.
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santhoshn
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#112
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Only iff x is an even integer
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iyappana
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#113
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Only iff x is even integer
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ashwinmeena
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#114
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Only if x is even integer
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