IMO 2024 · Problem 1

Determine all real numbers \(\alpha\) such that, for every positive integer \(n\), the integer \[ \lfloor\alpha\rfloor+\lfloor2\alpha\rfloor+\cdots+\lfloor n\alpha\rfloor \] 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 \(\lfloor2\rfloor=\lfloor2.9\rfloor=2\).)

Official text of the problem. Official problem paper (PDF)

Answer

All even integers: \[ \boxed{\alpha\in2\mathbb Z}. \]

Solution

Suppose \(\alpha\) satisfies the condition. For every positive integer \(n\), define \[ A_n=\frac1n\sum_{k=1}^{n}\lfloor k\alpha\rfloor. \] By hypothesis, every \(A_n\) is an integer.

For \(n\ge2\), summing the inequalities \[ k\alpha-1<\lfloor k\alpha\rfloor\le k\alpha \] over \(1\le k\le n-1\), then dividing by \(n-1\), gives \[ \frac{n\alpha}{2}-1<A_{n-1}\le\frac{n\alpha}{2}. \] Also, \[ n\alpha-1<\lfloor n\alpha\rfloor\le n\alpha. \] Since \[ n(A_n-A_{n-1})=\lfloor n\alpha\rfloor-A_{n-1}, \] these bounds imply \[ \frac{\alpha}{2}-\frac1n <A_n-A_{n-1}< \frac{\alpha}{2}+\frac1n. \] Consequently, the integer sequence \(A_n-A_{n-1}\) converges to \(\alpha/2\).

Every convergent sequence of integers is eventually constant: all sufficiently late terms lie in an interval of length less than \(1\), which contains at most one integer. Its limit is therefore an integer. Hence \(\alpha/2\in\mathbb Z\), so \(\alpha\) is an even integer.

Conversely, if \(\alpha=2m\) for some \(m\in\mathbb Z\), then for every positive integer \(n\), \[ \sum_{k=1}^{n}\lfloor k\alpha\rfloor =\sum_{k=1}^{n}2mk =mn(n+1), \] which is a multiple of \(n\). Thus every even integer satisfies the condition.

The solution was checked against the official IMO shortlist solutions.