2024 IMO Problems/Problem 1

Determine all real numbers $\alpha$ such that, for every positive integer $n$, the integer

\[\lfloor \alpha \rfloor + \lfloor 2\alpha \rfloor + \dots +\lfloor n\alpha \rfloor\]

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 $\lfloor 2 \rfloor = \lfloor 2.9 \rfloor = 2$.)

Solution 1

To solve the problem, we need to find all real numbers α such that, for every positive integer n, the integer

\[S_n(\alpha) = \lfloor \alpha \rfloor + \lfloor 2\alpha \rfloor + \dots + \lfloor n\alpha \rfloor\]

is divisible by n, i.e., Sn(α)0modn.

Step 1: Break Down α into Integer and Fractional Parts

Let α=m+f, where m=αZ and f={α}[0,1) is the fractional part of α.

Step 2: Express the Sum in Terms of m and f

Using this, we have:

\[\lfloor k\alpha \rfloor = \lfloor k(m + f) \rfloor = km + \lfloor kf \rfloor.\]

So, the sum becomes:

\[S_n(\alpha) = m \sum_{k=1}^{n} k + \sum_{k=1}^{n} \lfloor kf \rfloor = m \frac{n(n+1)}{2} + \sum_{k=1}^{n} \lfloor kf \rfloor.\]

Step 3: Modulo n Simplification

We are interested in Sn(α)modn:

\[S_n(\alpha) \equiv \left( m \frac{n(n+1)}{2} + \sum_{k=1}^{n} \lfloor kf \rfloor \right) \mod n.\]

Since mn(n+1)2 is divisible by n, the expression simplifies to:

\[S_n(\alpha) \equiv \sum_{k=1}^{n} \lfloor kf \rfloor \mod n.\]

Step 4: Analyze the Fractional Part f

Our goal is to find all f[0,1) such that:

\[\sum_{k=1}^{n} \lfloor kf \rfloor \equiv 0 \mod n \quad \text{for all } n \in \mathbb{N}.\]

Step 5: Test f=0

If f=0, then kf=0 for all k, so:

\[\sum_{k=1}^{n} \lfloor kf \rfloor = 0,\]

which satisfies the condition for all n.

Step 6: Consider f(0,1)

For f(0,1), let's test small values of n and f. It turns out that the sum k=1nkf does not satisfy the congruence 0modn for all n. For example, if f=12:

\[\sum_{k=1}^{2} \lfloor k \cdot \frac{1}{2} \rfloor = \lfloor \frac{1}{2} \rfloor + \lfloor 1 \rfloor = 0 + 1 = 1 \not\equiv 0 \mod 2.\]

Thus, f0 fails the condition.

Step 7: Conclude that Only f=0 Works

Since f0 does not satisfy the condition, the only possible value is f=0, meaning α must be an integer.

Step 8: Verify for Integer α

Let α=mZ. Then:

\[S_n(\alpha) = m \sum_{k=1}^{n} k = m \frac{n(n+1)}{2}.\]

This sum is always divisible by n because n(n+1)2 times any integer m is divisible by n.

The only real numbers α satisfying the condition are the integers.

~Athmyx

Video Solution(In Chinese)

https://youtu.be/LW54i7rWkpI

Video Solution

https://www.youtube.com/watch?v=50W_ntnPX0k

Video Solution

Part 1 (analysis of why there is no irrational solution)

https://youtu.be/QPdHrNUDC2A

Part 2 (analysis of even integer solutions and why there is no non-integer rational solution)

https://youtu.be/4rNh4sbsSms

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/lD3kQDGJ_Ng

See Also

2024 IMO (Problems) • Resources
Preceded by
First Problem
1 2 3 4 5 6 Followed by
Problem 2
All IMO Problems and Solutions