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 $ \alpha $ 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(\alpha) \equiv 0 \mod n $.

Step 1: Break Down $ \alpha $ into Integer and Fractional Parts

Let $ \alpha = m + f $, where $ m = \lfloor \alpha \rfloor \in \mathbb{Z} $ and $ f = \{\alpha\} \in [0,1) $ is the fractional part of $ \alpha $.

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(\alpha) \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 $ \lfloor kf \rfloor = 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} \lfloor kf \rfloor $ 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 \ne 0 $ fails the condition.

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

Since $ f \ne 0 $ does not satisfy the condition, the only possible value is $ f = 0 $, meaning $ \alpha $ must be an integer.

Step 8: Verify for Integer $ \alpha $

Let $ \alpha = m \in \mathbb{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 $ \alpha $ 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

Video Solution

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