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., $S_{n} (α) \equiv 0 \mod n$ .

Step 1: Break Down $α$ into Integer and Fractional Parts

Let $α = m + f$ , where $m = ⌊ α ⌋ \in Z$ and $f = {α} \in [0, 1)$ 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 $S_{n} (α) \mod n$ :

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

Since $m \frac{n (n + 1)}{2}$ 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 $f \in [0, 1)$ 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 $⌊ k f ⌋ = 0$ for all $k$ , so:

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

which satisfies the condition for all $n$ .

Step 6: Consider $f \in (0, 1)$

For $f \in (0, 1)$ , let's test small values of $n$ and $f$ . It turns out that the sum $\sum_{k = 1}^{n} ⌊ k f ⌋$ does not satisfy the congruence $0 \mod n$ for all $n$ . For example, if $f = \frac{1}{2}$ :

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

Thus, $f \neq 0$ fails the condition.

Step 7: Conclude that Only $f = 0$ Works

Since $f \neq 0$ 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 $α = m \in Z$ . Then:

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

This sum is divisible by $n$ if and only if $m$ is even.

Because $\frac{n (n + 1)}{2}$ times any even integer $m$ is divisible by $n$ .

But $\frac{n (n + 1)}{2}$ times an odd integer $m$ is not, if n is even.

The only real numbers $α$ satisfying the condition are the even 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

2024 IMO (Problems) • Resources
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2024 IMO Problems/Problem 1

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Solution 1

Video Solution(In Chinese)

Video Solution

Video Solution

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