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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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0 replies
jlacosta
May 1, 2025
0 replies
Another NT FE
nukelauncher   63
N a minute ago by sansgankrsngupta
Source: ISL 2019 N4
Find all functions $f:\mathbb Z_{>0}\to \mathbb Z_{>0}$ such that $a+f(b)$ divides $a^2+bf(a)$ for all positive integers $a$ and $b$ with $a+b>2019$.
63 replies
nukelauncher
Sep 22, 2020
sansgankrsngupta
a minute ago
strange geometry problem
Zavyk09   1
N 7 minutes ago by Captainscrubz
Source: own
Let $ABC$ be a triangle with circumcenter $O$ and internal bisector $AD$. Let $AD$ cuts $(O)$ again at $M$ and $MO$ cuts $(O)$ again at $N$. Point $L$ lie on $AD$ such that $(AD, LM) = -1$. The line pass through $L$ and perpendicular to $AD$ intersects $NC, NB$ at $P, Q$ respectively. Let circumcircle of $\triangle NPQ$ cuts $(O)$ at $G \ne N$. Prove that $\angle AGD = 90^{\circ}$.
1 reply
Zavyk09
Yesterday at 4:32 PM
Captainscrubz
7 minutes ago
n^6 + 5n^3 + 4n + 116 is the product of two or more consecutive numbers
Amir Hossein   2
N 23 minutes ago by KTYC
Source: Bulgaria JBMO TST 2018, Day 1, Problem 3
Find all positive integers $n$ such that the number
$$n^6 + 5n^3 + 4n + 116$$is the product of two or more consecutive numbers.
2 replies
Amir Hossein
Jun 25, 2018
KTYC
23 minutes ago
IMO Shortlist 2009 - Problem G3
April   49
N an hour ago by Ilikeminecraft
Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram.
Prove that $GR=GS$.

Proposed by Hossein Karke Abadi, Iran
49 replies
April
Jul 5, 2010
Ilikeminecraft
an hour ago
No more topics!
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
Y by
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
575 posts
#110
Y by
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
8868 posts
#111
Y by
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
5 posts
#112
Y by
Only iff x is an even integer
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iyappana
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#113
Y by
Only iff x is even integer
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ashwinmeena
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#114
Y by
Only if x is even integer
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