Very nice problem! I think that answer is all even $\alpha$. They are clearly works. Now let $2 \not | \alpha$. WLOG $0<\alpha<2$, since we can translate it on $2$. Let $$\frac{\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor}{n}=f(n)/n.$$Then $\frac{f(n+1)}{n+1}-\frac{f(n)}{n}=\frac{n\lfloor{\alpha(n+1)\rfloor}-f(n)}{n(n+1)}$ is integer and so $n(n+1)|n\lfloor\alpha(n+1)\rfloor-f(n)$. But $0<^*RHS<2n(n+1)$ and so $g(n)=RHS=n(n+1)$. But now we can look at $g(n+1)$ and get $\lfloor{\alpha(n+2) \rfloor}=2+\lfloor (n+1)\alpha \rfloor$ for all $n$, which is impossible for $0<\alpha <2$. (We can use induction on $n$ to get $\lfloor \alpha (n+1) \rfloor =2n$ and thus $\frac{2n-1}{n+1}< \alpha <\frac{2n+1}{n+1}$ for all $n$). (*) Here we used the fact that $\alpha>1$. In other case $0<g(n)<n(n+1)$ for large $n$, which is impossible.

Answer is all even integers. If $\alpha=2k,$ then $\frac{n(n+1)}{2}.2k=n(n+1)k$ is divisible by $n$. Let's show that other real numbers does not satisfy the condition. If $\alpha$ satisfies, then $\alpha+2$ also satisfies hence we can assume that $0\leq \alpha<2$. Let $\alpha\neq 0\iff 0<\alpha<2$. If $\alpha<1,$ denote $m$ as $\lceil \frac{1}{\alpha}\rceil=m\iff (m-1)\alpha<1\leq m\alpha$ where $2\leq m\in Z^+$. Since $m|\lfloor \alpha \rfloor+...+\lfloor (m-1)\alpha \rfloor+\lfloor m\alpha \rfloor=\lfloor m\alpha \rfloor$ But $1\leq \lfloor m\alpha \rfloor<m$ which is impossible. Also $\alpha=1$ doesn't hold which can be seen by taking $n=2$. Thus, $1<\alpha<2$. Let $\alpha=1+\beta$ where $0<\beta<1$. We have $n|\frac{n(n+1)}{2}+\sum{\lfloor i\beta} \rfloor$ Let $\lceil \frac{1}{\beta}\rceil =k$ We have $k|\frac{k(k+1)}{2}+\lfloor k\beta \rfloor=\frac{k(k+1)}{2}+1$ since $(k-1)\beta<1$ yields $k\beta<\beta+1<2$. So $2k|k^2+k+2\implies k=1,2$. $\frac{1}{\beta}>1\implies k=2$. Let's show that if $\lfloor m\beta \rfloor=m-1$ for all $1\leq m\leq N,$ then $\lfloor (m+1)\beta \rfloor=m$ for $m\geq 3$. \[\frac{m+1}{2}|m+1|\frac{(m+1)(m+2)}{2}+\frac{(m-1)m}{2}+\lfloor (m+1)\beta \rfloor\equiv \frac{m^2-m+m^2+3m+2}{2}+\lfloor (m+1)\beta \rfloor\equiv 1+\lfloor (m+1)\beta \rfloor\]Since $m+1|2+2\lfloor (m+1)\beta \rfloor$ and $m-1\leq \lfloor (m+1)\beta \rfloor<m+1,$ we get $\lfloor (m+1)\beta \rfloor=m$ which completes our induction. So $\lfloor n\beta \rfloor=n-1$ for all $n$ but this is impossible since it gives that $n\beta>n-1\iff \beta>\frac{n-1}{n}=1-\frac{1}{n}$ where $n$ goes to infinity. Therefore $\alpha \not \in (0,2)$ as desired.$\blacksquare$

I’m really excited to have a P1 on IMO for the second year in a row. This problem was proposed by me Santiago Rodríguez from Colombia. I genuinely hope that you enjoy my problem as much as I do.

@above Thanks!!! It's a beautiful problem!!

$\alpha$ any even integer works, because $2(1+2+\cdots+n)=n(n+1)$. We claim these are the only solutions. If $\alpha$ is not an even integer, we can add any even integer to $\alpha$ without changing whether the condition holds (due to the same identity), so we may assume that $-1<\alpha\le1$. Case 1: $\alpha=1$. This fails because $1+2=3$ is not a multiple of 2. Case 2: $0<\alpha<1$. Let $k>1$ be the smallest integer such that $k\alpha\ge1$; note that $k\alpha=(k-1)\alpha+\alpha<1+1=2$. Hence taking $n=k$, the sum is $0+0+\cdots+0+1=1$, which is not a multiple of $n$. Case 3: $-1<\alpha<0$. Let $k>1$ be the smallest integer such that $k\alpha<-1$; note that $k\alpha=(k-1)\alpha+\alpha>(-1)+(-1)=-2$. Hence taking $n=k$, the sum is $(-1)+(-1)+\cdots+(-1)+(-2)=-n-1$, which is not a multiple of $n$. Hence the only $\alpha$ which work are the even integers.

Floor functions are becoming popular in international contests nowadays. It's interesting. Nice problem @Supertinito.

Let $\alpha = a + b$ where $a = \lfloor \alpha \rfloor$ and $b = \alpha - \lfloor \alpha \rfloor$. Then we get that $n$ divides $$\frac{a}{2}(n(n + 1)) + \lfloor 2b \rfloor + \dots \lfloor nb \rfloor$$ Then if $a$ is even we have $n \mid \sum_{m = 2}^n \lfloor mb \rfloor$ from here we see clearly that $\lfloor mb \rfloor = 0$ for all $m$. This of course implies $b = 0$ therefore $\alpha = a = 2k$. Now suppose $a$ is odd. Then we have that $\lfloor 2b \rfloor = 1$ and $n \mid \sum_{m = 2}^n \lfloor mb \rfloor$ for all odd $n$. From this inductively we get $mb \geq \lfloor mb \rfloor = m - 1$ a contradiction as we know $b < 1$ on RHS is linear function of higher slope.

Notice that $\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor$ is the number of integer points in [1,n] within the interval (excluding boundaries) between y=nx and y=0

What's the genre of the problem? Is it A or N?

shanelin-sigma wrote: What's the genre of the problem? Is it A or N? It’s algebra as P2 is NT

Funcshun840 wrote: It’s algebra as P2 is NT But I think this problem is more like an NT problem since it reminds me some steps when proving quadratic reciprocity Besides I think P3 is also an NT problem NNN this year?

shanelin-sigma wrote: Funcshun840 wrote: It’s algebra as P2 is NT But I think this problem is more like an NT problem since it reminds me some steps when proving quadratic reciprocity Besides I think P3 is also an NT problem NNN this year? If 2020 IMO P6 can be a G, I don't see how this can't be an A Also P3 clearly is C, I don't see any number theory elements in it.

CrazyInMath wrote: If 2020 IMO P6 can be a G, I don't see how this can't be an A : D CrazyInMath wrote: Also P3 clearly is C, I don't see any number theory elements in it. QAQ

NO_SQUARES wrote: Very nice problem! Let $$\frac{\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor}{n}=f(n)/n.$$Then $f(n+1)-f(n)=\frac{n\lfloor{\alpha(n+1)\rfloor}-f(n)}{n(n+1)}$ is integer areyou sure is that the definition of $f$? I thought it has typos (and it is very confusing at the beginning)

This was pretty straightforward but grinding out the details was slightly messy. Hope I managed to completely nail this problem. We claim that the answer is all even integers. It is easy to see that these indeed work since for all $\alpha = 2r$ for $r\in \mathbb{Z}$, we have \[\lfloor \alpha \rfloor + \lfloor 2\alpha \rfloor + \dots + \lfloor n\alpha \rfloor = 2r + 4r + \dots + 2nr = rn(n+1)\]which is quite clearly a multiple of $2r$ since at least one of $n$ and $n+1$ is a multiple of two. Now, we start off by solving the following simpler problem. Dealing with the fractional parts wrote: Determine all positive real numbers $0\le \epsilon < 1$ for which \[2(\lfloor \epsilon \rfloor + \lfloor 2\epsilon \rfloor + \dots + \lfloor n \epsilon \rfloor) \]is a multiple of $n$ for all positive integers $n$. It is clear that when $\epsilon =0$ the desired characteristic is satisfied so we deal with $0< \epsilon < 1$. Now, let $\frac{1}{k} > \epsilon \ge \frac{1}{k+1}$ for some positive integer $k$. Then, for all $n_0 \le k$, \[\lfloor n_0 \epsilon \rfloor< \lfloor \frac{n_0}{k} \rfloor \le \lfloor \frac{k}{k} \rfloor = 1\]from which it is clear that $\lfloor n_0 \epsilon \rfloor = 0$. Further, \[\lfloor (k+1) \epsilon \rfloor < \lfloor\frac{k+1}{k} \rfloor \le 2\]for all $k \in \mathbb{N}$. Thus, $ \lfloor (k+1) \epsilon \rfloor =1$. But then, \[k+1 \mid 2(\lfloor \epsilon \rfloor + \lfloor 2\epsilon \rfloor + \dots + \lfloor n \epsilon \rfloor) = 2(1) = 2\]which requires $k=1$. This implies $\frac{1}{2} \le \epsilon < 1$. But then, we can prove the following claim via induction. Claim : For all positive $\epsilon$ which satisfy the desired property we must have $\lfloor n\epsilon\rfloor = n-1$ for all positive integers $n$. Proof : The claim is clear when $n=1$ and $n=2$. To see why it holds for all other $n$ say it holds for some $m >1$. Then, \[m+1 \mid 2(\lfloor \epsilon \rfloor + \lfloor 2\epsilon \rfloor + \dots + \lfloor m \epsilon \rfloor + \lfloor (m+1)\epsilon \rfloor) = (m-1)m +2\lfloor (m+1) \epsilon \rfloor\]So, \begin{align*} m^2 -m + 2\lfloor (m+1)\epsilon \rfloor & \equiv 0 \pmod{m+1}\\ 2\lfloor (m+1)\epsilon \rfloor & \equiv 2m \pmod{m+1} \end{align*}Now, we split into two cases. If $m$ is even, this implies $\lfloor (m+1)\epsilon \rfloor \equiv m \pmod{m+1}$. Further, $\lfloor(m+1)\epsilon \rfloor > 0$ and $\lfloor (m+1)\epsilon \rfloor < m+1$ so we must have $\lfloor (m+1)\epsilon \rfloor =m$. If $m$ is odd, this implies $\lfloor (m+1)\epsilon \rfloor \equiv m \pmod{\frac{m+1}{2}}$. Then, since we know $\epsilon \ge \frac{1}{2}$, we know $\lfloor (m+1)\epsilon \rfloor \ge \lfloor \frac{m+1}{2} \rfloor = \frac{m+1}{2}$. Further, $\lfloor (m+1)\epsilon \rfloor < \lfloor m+1 \rfloor = m+1$, which then again implies that $\lfloor (m+1)\epsilon \rfloor=m$ as desired. So, the induction is complete and our claim is proved. But now, it is quite clear that the claim cannot hold for all $n \in \mathbb{N}$ since if we consider $\frac{k}{k+1} \ge \epsilon > \frac{k+1}{k+2}$, and look at $\lfloor (k+2)\epsilon \rfloor < \lfloor k+1 \rfloor = k+1 $ which contradicts our claim. Thus, no $\epsilon >0$ which satisfies the desired problem exist. In other words, if there exists such $\alpha$ they must be integers. It is further, clear that $\alpha$ must be an even integer since if $\alpha$ is an odd integer, \[2\mid \lfloor \alpha \rfloor + \lfloor 2\alpha \rfloor =3\alpha\]which is a clear contradiction for odd $\alpha$. Thus, all solutions are of the claim types and we are done.

PfTB-style bounding The answer is all even integers. If $\alpha=2k$ for some integers $k$, then $$\lfloor\alpha\rfloor + \lfloor 2\alpha\rfloor + \dots + \lfloor n\alpha\rfloor = 2k\cdot \frac{n(n+1)}{2} = nk(n+1),$$which works. Now, we prove that $\alpha$ must be an even integer. To do that, define $$x_n = \frac{\lfloor\alpha\rfloor + \lfloor 2\alpha\rfloor + \dots + \lfloor n\alpha\rfloor}{n},$$so that \begin{align*} x_{n+1}-x_n &= \frac{\lfloor\alpha\rfloor + \lfloor 2\alpha\rfloor + \dots + \lfloor (n+1)\alpha\rfloor}{n+1} - \frac{\lfloor\alpha\rfloor + \lfloor 2\alpha\rfloor + \dots + \lfloor n\alpha\rfloor}{n} \\ &= \frac{\lfloor (n+1)\alpha\rfloor}{n+1} - \frac{\lfloor\alpha\rfloor + \lfloor 2\alpha\rfloor + \dots + \lfloor n\alpha\rfloor}{n(n+1)} \\ &= \frac{\alpha(n+1)+O(1)}{n+1} - \frac{\alpha\cdot \tfrac{n(n+1)}{2} + O(n)}{n(n+1)} \\ &= \frac{\alpha}{2} + O\left(\frac 1n\right), \end{align*}so $x_{n+1}-x_n \to \tfrac{\alpha}2$, and since each term is an integer, it follows that the sequence is eventually constant and $\tfrac\alpha 2\in\mathbb Z$.

EthanWYX2009 wrote: Find all real number $\alpha,$ such that for any positive integer $n,$ $$\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor$$is a multiple of $n.$ Does this work? Call the sum $S(a, n)$, $\alpha$ replaced by $a$. Translate $a$ to be in $(0, 2)$. For $a \in (0, 1)$, take $n=\left \lceil \dfrac{1}{a} \right \rceil$ to get $n \lvert 1$. $a \in \left [1, 2\right )$, if $2-\frac{1}{k-1}\leq a < 2-\frac{1}{k}$ for positive integer $k$, then $n=k \implies S(a, n) = 1+3+\dots 2k-3+2k-2=k^2-1$ not divisible by $k$. Please inform me about typos.

Let $f_{k}(\alpha) =\lfloor{k\alpha}\rfloor$ and $N = \sum_{k=1}^n f_k(\alpha)$. If $\alpha$ is a solution to the problem, so is $\alpha + 2m$, where $m$ is an arbitrary integer. This is because $\lfloor{k(\alpha + 2m)}\rfloor = 2km + \lfloor{k\alpha}\rfloor$ and if $n | N$ then $n | \sum_{k=1}^n f_k(\alpha + 2m) = 2km + \sum_{k=1}^n \lfloor{k\alpha}\rfloor = N + \frac{2mn(n+1)}{2} = N + mn(n+1)$. This means we can limit our findings on an arbitrary interval of length 2 and extrapolate from there. For sake of simplicity take the $\alpha \in [-1,1)$. We split the problem into two cases: Case 1: Interval [-1,0) This means $-1\leq \alpha < 0 \implies -n \leq n\alpha < 0 \implies \lfloor{n\alpha}\rfloor \in \{-n,-n+1,\dots,-2,-1\}$ Claim: $\lfloor{n\alpha}\rfloor = -1$ for all positive integers n. Using strong induction, base case is trivial $1 | \lfloor{\alpha}\rfloor = -1$. Assume $\lfloor{n\alpha}\rfloor = -1$ for all $n \in \{1,2,\dots,k-1,k \}$ Consider $n = k+1$: $n = k+1 | -k + \lfloor{(k+1)\alpha}\rfloor$. Now $ \lfloor{(k+1)\alpha}\rfloor \in \{-(k+1),-k,\dots,-2,-1\}$ The only way for $k+1$ to divide $-k + \lfloor{(k+1)\alpha}\rfloor$ is for $\lfloor{(k+1)\alpha}\rfloor = -1$. This concludes the proof. Now, since $\lfloor{n\alpha}\rfloor = -1$, $-1\leq n\alpha < 0$. Thus $-\frac{1}{n}\leq \alpha < 0$. As the problem's condition holds true for any positive integer n, it must hold for arbitrarily large n, i.e. on the limiting case as n tends to infinity. Taking the limit on our inequality yields $0\leq \alpha < 0$, with no $\alpha$ satisfying the double inequality. Case 2: Interval [0,1) This means $0\leq \alpha < 1 \implies 0 \leq n\alpha < n \implies \lfloor{n\alpha}\rfloor \in \{0,1,\dots,n-2,n-1\}$. Claim: $\lfloor{n\alpha}\rfloor = 0$ for all positive integers n. Using strong induction, base case is again trivial $1 | \lfloor{\alpha}\rfloor = 0$. Assume $\lfloor{n\alpha}\rfloor = 0$ for all $n \in \{1,2,\dots,k-1\}$ Consider $n = k+1$: $n = k+1 | 0 + \lfloor{(k+1)\alpha}\rfloor$. Now $\lfloor{(k+1)\alpha}\rfloor \in \{0,1,2,\dots,k-1,k\}$ The only way for $k+1$ to divide $\lfloor{(k+1)\alpha}\rfloor$ is for $\lfloor{(k+1)\alpha}\rfloor = 0$. This concludes the proof. Since $\lfloor{n\alpha}\rfloor = 0$ for any n, it implies $\alpha = 0$. Thus the only reals $\alpha$ satisfying the problem are $\alpha = 2m$, where m is an integer.

Let $\alpha=m+x$ where $m$ is an integer and $x\in [0,1)$. Define $s_n:=2\cdot \sum_{k=1}^n \lfloor kx \rfloor$. I'll say $x$ works if for all $n\mid s_n$ for all $n$. Assume $x>0$. Suppose $x=p/q$ where $p<q$ and they are coprime positive integers. Note that $s_{q-1}=(p-1)(q-1)$. So, $s_q=(p-1)(q-1)+2p \equiv p+1 \pmod q$. Thus since $p<q$ we must have $p+1=q$. Since $\left \lfloor \frac{p(q+1)}q \right \rfloor =\left \lfloor \frac{q^2-1}q\right \rfloor=q-1$, we have $$s_{q+1}=(q-2)(q-1)+4(q-1)\equiv (-3)(-2)+4(-2)\equiv -2 \pmod{q+1}.$$Thus rational $x$ doesn't work. If $x$ were real then we could find a rational $0<r<1$ arbitrarily close to $x$ since $\mathbb Q$ is dense in $\mathbb R$. And in that case, for any fixed integer $n$, $\lfloor kx \rfloor =\lfloor kr\rfloor$ for $1\leq k\leq n$ would hold. Thus since $r$ doesn't work, $x$ wouldn't either. Now, notice that twice sum in the problem would be $m\cdot n(n+1)+s_n$ and this (as we've seen above) isn't divisible by $n$ for every $n$ when $x>0$. Thus we must consider $x=0$. In this case, clearly even integers only satisfy the problem.

pie854 wrote: Let $\alpha=m+x$ where $m$ is an integer and $x\in [0,1)$. Define $s_n:=2\cdot \sum_{k=1}^n \lfloor kx \rfloor$. I'll say $x$ works if for all $n\mid s_n$ for all $n$. Assume $x>0$. Suppose $x=p/q$ where $p<q$ and they are coprime positive integers. Note that $s_{q-1}=(p-1)(q-1)$. So, $s_q=(p-1)(q-1)+2p \equiv p+1 \pmod q$. Thus since $p<q$ we must have $p+1=q$. Since $\left \lfloor \frac{p(q+1)}q \right \rfloor =\left \lfloor \frac{q^2-1}q\right \rfloor=q-1$, we have $$s_{q+1}=(q-2)(q-1)+4(q-1)\equiv (-3)(-2)+4(-2)\equiv -2 \pmod{q+1}.$$Thus rational $x$ doesn't work. If $x$ were real then we could find a rational $0<r<1$ arbitrarily close to $x$ since $\mathbb Q$ is dense in $\mathbb R$. And in that case, for any fixed integer $n$, $\lfloor kx \rfloor =\lfloor kr\rfloor$ for $1\leq k\leq n$ would hold. Thus since $r$ doesn't work, $x$ wouldn't either. Now, notice that twice sum in the problem would be $m\cdot n(n+1)+s_n$ and this (as we've seen above) isn't divisible by $n$ for every $n$ when $x>0$. Thus we must consider $x=0$. In this case, clearly even integers only satisfy the problem. I don't understand the part , why $p<q$ and $s_q\equiv p+1\pmod{q}$ implies, $p+1=q$ ? Can you please clarify? Also , dangerousliri wrote: This is the solution I and Leader of USA (John Berman) found during the process of selecting the test. Let's write with $b_n=\frac{\lfloor \alpha\rfloor+\lfloor 2\alpha\rfloor+...+\lfloor n\alpha\rfloor}{n}$ which is an integer from the condition. Since it is well known $x-1<\lfloor x\rfloor\leq x$ we have, $$\frac{n+1}{2}\alpha-1<b_n\leq \frac{n+1}{2}\alpha.$$ And because $b_n$ is an integer we have that $b_n=\left\lfloor\frac{n+1}{2}\alpha\right\rfloor$. So, $$\lfloor n\alpha\rfloor=nb_n-(n-1)b_{n-1}=n\left\lfloor\frac{n+1}{2}\alpha\right\rfloor-(n-1)\left\lfloor\frac{n}{2}\alpha\right\rfloor$$ Taking modulo $n$ in last equation we have, $$\lfloor n\alpha\rfloor\equiv\left\lfloor\frac{n}{2}\alpha\right\rfloor\pmod{n}$$ Using $n\alpha-1<\lfloor n\alpha\rfloor\leq n\alpha$ we have, $\frac{\lfloor n\alpha\rfloor\pmod{n}}{n}$ approaches $\{\alpha\}$ as $n\rightarrow\infty$. Hence the above implies, $$\left\{\alpha\right\}=\left\{\frac{\alpha}{2}\right\},$$ which means $\frac{\alpha}{2}=\alpha-\frac{\alpha}{2}\in\mathbb{Z}$ so $\alpha$ is an even integer. And we can easily check that all even integers $\alpha$ works. Superb solution. In bengali we call it aladai.

Answer: $\alpha=2r, r\in\mathbb{Z}$ $S_{n}=\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor$ $\frac{S_{n}}{n}-\frac{S_{n+1}}{n+1}=\frac{\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor}{n(n+1)}-\frac{\lfloor (n+1)\alpha\rfloor}{n+1}\Rightarrow\lim\limits_{n\to\infty}(\frac{S_{n}}{n}-\frac{S_{n+1}}{n+1}) =\frac{\alpha}{2}-\alpha=-\frac{\alpha}{2}$ Because $\frac{S_{n}}{n}-\frac{S_{n+1}}{n+1}\in\mathbb{Z}$, we get $\frac{\alpha}{2}\in\mathbb{Z}\Rightarrow\alpha$ is even. Clearly $n\mid \alpha\cdot\frac{n(n+1)}{2}$ for even $\alpha$, proving our claim.

Safal wrote: I don't understand the part , why $p<q$ and $s_q\equiv p+1\pmod{q}$ implies, $p+1=q$ ? Can you please clarify? The condition is that $$m\cdot q(q+1)+s_q\equiv 0\pmod q$$so $p+1\equiv s_q\equiv 0 \pmod q$. This means $p+1\geq q$. Also $p<q \implies p+1\leq q$. So $p+1=q$. Also West Bengali or Bangladeshi?

A few things, I edited my original post (#5) to add the story of the creation of the problem (because people seem to like that kind of thing and I wanted to be like the other cool problem authors ). Also, I wanted to share the following solution by Felix Hamoen (NLD 2)

I find it quite interesting that it only uses the divisibility relation from the problem statement for $n$ of the form $n=2^\ell$. This sparked my interest in the following generalization of the problem: Let $A$ be some subset of the positive integers. Let $\alpha$ be a real number such that, for every positive integer $n\in A$ the integer $$\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor$$is a multiple of $n.$ For which $A$ does it hold that $\alpha$ has to be an even integer? The solution that I just shared shows that if $A$ is the set of the powers of 2 then it works for the generalized problem. The "limit-taking solutions" (such as the one in post #19) work if $A$ contains infinitely many pairs of consecutive integers. I find it weird how these two kinds of sets feel so different but still satisfy the condition. It is not hard to find counterexamples for finite $A$. If anyone makes some progress towards this version of the problem, please let me know!

A bit easy for IMO P1. Motivation : If any $ \alpha \in R $ satisfy the equation $\iff$ $\alpha -2 $ satisfy the equation. Proof : Define a function say $T(\alpha)$ = $\lfloor {\alpha}\rfloor + \lfloor {2\alpha} \rfloor + . . . + \lfloor {n\alpha} \rfloor $. Then, $T(\alpha-2) = \lfloor {\alpha-2} \rfloor + \lfloor {2 (\alpha-2)} \rfloor + ... + \lfloor {n (\alpha - 2)} \rfloor $. We can say that $T(\alpha) - T(\alpha - 2) = 2 (\frac{n(n+1)}{2})= n(n+1) $ is a multiple of $n$.

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.

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. $\blacksquare$

We uploaded our solution https://calimath.org/pdf/IMO2024-1.pdf on youtube https://youtu.be/ewvE5_kNw1I.

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.

\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}

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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