IMO2024 Shortlisted Problems

Answer: All even integers satisfy the condition of the problem and no other real number $\alpha$ does so.

Solution 1. First we will show that even integers satisfy the condition. If $\alpha =2m$ where $m$ is an integer then $$\lfloor \alpha \rfloor +\lfloor 2\alpha \rfloor +\cdots +\lfloor n\alpha \rfloor =2m+4m+\cdots +2mn=mn(n+1)$$ which is a multiple of $n$.

Now we will show that they are the only real numbers satisfying the conditions of the problem. Let $\alpha =k+ϵ$ where $k$ is an integer and $0⩽ϵ<1$. Then the number $$\begin{array}{rl}\lfloor \alpha \rfloor +\lfloor 2\alpha \rfloor +\cdots +\lfloor n\alpha \rfloor & =k+\lfloor ϵ\rfloor +2k+\lfloor 2ϵ\rfloor +\cdots +nk+\lfloor nϵ\rfloor \\ & =\frac{kn(n+1)}{2}+\lfloor ϵ\rfloor +\lfloor 2ϵ\rfloor +\cdots +\lfloor nϵ\rfloor \end{array}$$ has to be a multiple of $n$. We consider two cases based on the parity of $k$.

Case 1: $k$ is even. Then $\frac{kn(n+1)}{2}$ is always a multiple of $n$. Thus $$\lfloor ϵ\rfloor +\lfloor 2ϵ\rfloor +\cdots +\lfloor nϵ\rfloor$$ also has to be a multiple of $n$.

We will prove that $\lfloor nϵ\rfloor =0$ for every positive integer $n$ by strong induction. The base case $n=1$ follows from the fact that $0⩽ϵ<1$. Let us suppose that $\lfloor mϵ\rfloor =0$ for every $1⩽m<n$. Then the number $$\lfloor ϵ\rfloor +\lfloor 2ϵ\rfloor +\cdots +\lfloor nϵ\rfloor =\lfloor nϵ\rfloor$$ has to be a multiple of $n$. As $0⩽ϵ<1$ then $0⩽nϵ<n$, which means that the number $\lfloor nϵ\rfloor$ has to be equal to 0.

The equality $\lfloor nϵ\rfloor =0$ implies $0⩽ϵ<1/n$. Since this has to happen for all $n$, we conclude that $ϵ=0$ and then $\alpha$ is an even integer.

Case 2: $k$ is odd. We will prove that $\lfloor nϵ\rfloor =n-1$ for every natural number $n$ by strong induction. The base case $n=1$ again follows from the fact that $0⩽ϵ<1$. Let us suppose that $\lfloor mϵ\rfloor =m-1$ for every $1⩽m<n$. We need the number $$1+\lfloor nϵ\rfloor$$ to be a multiple of $n$. As $k$ is odd, we need $1+\lfloor nϵ\rfloor$ to be a multiple of $n$. Again, as $0⩽ϵ<1$ then $0⩽nϵ<n$, so $\lfloor nϵ\rfloor =n-1$ as we wanted.

This implies that $1-\frac{1}{n}⩽ϵ<1$ for all $n$ which is absurd. So there are no other solutions in this case.

Comment. An alternative ending to the previous solution is as follows. By definition we have $S_n⩽\alpha \frac{n(n+1)}{2}$, on the other hand (5) implies $S_n⩾\alpha n^2-n$ for all $n$ large enough, so $\alpha =0$.

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Alternative solution.

As in Solution 1 we check that for even integers the condition is satisfied. Then, without loss of generality we can assume $0⩽\alpha <2$. We set $S_n=\lfloor \alpha \rfloor +\lfloor 2\alpha \rfloor +\cdots +\lfloor n\alpha \rfloor$.

Notice that $$\begin{array}{ll}{S}_n\equiv 0& (\mathrm{mod}n)\\ S_n\equiv S_n-S_{n-1}=\lfloor n\alpha \rfloor & (\mathrm{mod}n-1)\end{array}\text{\hspace{1em}(2)}$$ Since $\gcd (n,n-1)=1$, (1) and (2) imply that $$\begin{array}{c}{S}_n\equiv n\lfloor n\alpha \rfloor (\mathrm{mod}n(n-1)).\end{array}\text{\hspace{1em}(3)}$$ In addition, $$\begin{array}{c}0⩽n\lfloor n\alpha \rfloor -S_n=\sum _{k=1}^n(\lfloor n\alpha \rfloor -\lfloor k\alpha \rfloor )<\sum _{k=1}^n(n\alpha -k\alpha +1)=\frac{n(n-1)}{2}\alpha +n.\end{array}\text{\hspace{1em}(4)}$$ For $n$ large enough, the RHS of (4) is less than $n(n-1)$. Then (3) forces $$\begin{array}{c}0=S_n-n\lfloor n\alpha \rfloor =\sum _{k=1}^n(\lfloor n\alpha \rfloor -\lfloor k\alpha \rfloor )\end{array}\text{\hspace{1em}(5)}$$ for $n$ large enough.

Since $\lfloor n\alpha \rfloor -\lfloor k\alpha \rfloor ⩾0$ for $1⩽k⩽n$, we get from (5) that, for all $n$ large enough, all these inequalities are equalities. In particular $\lfloor \alpha \rfloor =\lfloor n\alpha \rfloor$ for all $n$ large enough, which is absurd unless $\alpha =0$.

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Alternative solution.

As in other solutions, without loss of generality we may assume that $0⩽\alpha <2$. Even integers satisfy the condition, so we assume $0<\alpha <2$ and we will derive a contradiction.

By induction on $n$, we will simultaneously show that $$\begin{array}{c}\lfloor \alpha \rfloor +\lfloor 2\alpha \rfloor +\cdots +\lfloor n\alpha \rfloor =n^2\\ \text{ and }\frac{2n-1}{n}⩽\alpha <2.\end{array}\text{\hspace{1em}(6)(7)}$$ The base case is $n=1$ : If $\alpha <1$, consider $m=\lceil \frac{1}{\alpha }\rceil >1$, then $$\lfloor \alpha \rfloor +\lfloor 2\alpha \rfloor +\cdots +\lfloor m\alpha \rfloor =1$$ is not a multiple of $m$, so we deduce (7). Hence, $\lfloor \alpha \rfloor =1$ and (6) follows.

For the induction step: assume the induction hypothesis to be true for $n$, then by (7) $$2n+1-\frac{1}{n}⩽(n+1)\alpha <2n+2$$ Hence, $$n^2+2n⩽\lfloor \alpha \rfloor +\lfloor 2\alpha \rfloor +\cdots +\lfloor n\alpha \rfloor +\lfloor (n+1)\alpha \rfloor =n^2+\lfloor (n+1)\alpha \rfloor <n^2+2n+2.$$ So, necessarily $\lfloor (n+1)\alpha \rfloor =2n+1$ and $$\lfloor \alpha \rfloor +\lfloor 2\alpha \rfloor +\cdots +\lfloor n\alpha \rfloor +\lfloor (n+1)\alpha \rfloor =(n+1{)}^2$$ in order to obtain a multiple of $n+1$. These two equalities give (6) and (7) respectively.

Finally, we notice that condition (7) being true for all $n$ gives a contradiction.

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Alternative solution.

As in other solutions without loss of generality we will assume that $0<\alpha <2$ and derive a contradiction. For each $n$, we define $$b_n=\frac{\lfloor \alpha \rfloor +\lfloor 2\alpha \rfloor +\cdots +\lfloor n\alpha \rfloor }{n},$$ which is a nonnegative integer by the problem condition and our assumption. Note that $$\lfloor (n+1)\alpha \rfloor ⩾\lfloor \alpha \rfloor ,\lfloor 2\alpha \rfloor ,\dots ,\lfloor n\alpha \rfloor \text{ and }\lfloor (n+1)\alpha \rfloor >\lfloor \alpha \rfloor$$ for all $n>\frac{1}{\alpha }$. It follows that $b_{n+1}>b_n⟹b_{n+1}⩾b_n+1$ for $n>\frac{1}{\alpha }$. Thus, for all such $n$, $$b_n⩾n+C$$ where $C$ is a fixed integer. On the other hand, the definition of $b_n$ gives $$b_n=\frac{\lfloor \alpha \rfloor +\lfloor 2\alpha \rfloor +\cdots +\lfloor n\alpha \rfloor }{n}⩽\frac{\alpha +2\alpha +\cdots +n\alpha }{n}=\frac{\alpha }{2}(n+1),$$ which is a contradiction for sufficiently large $n$.