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International Mathematical Olympiad

Number theory Difficulty 7.9 National Olympiad, round 2 Find the answer

Find all real number α,\alpha, such that for any positive integer n,n,
⌊α⌋+⌊2α⌋+⋯+⌊nα⌋\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor
is a multiple of n.n.

*

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

To find all real numbers α\alpha such that for any positive integer nn, the expression

Sn=⌊α⌋+⌊2α⌋+⋯+⌊nα⌋ S_n = \lfloor \alpha \rfloor + \lfloor 2\alpha \rfloor + \cdots + \lfloor n\alpha \rfloor

is a multiple of nn, let's analyze the problem using properties of the floor function.

### Step 1: Analyze Sums of Floor Functions

For any α\alpha, we can express each floor term as:

⌊kα⌋=kα−{kα}, \lfloor k\alpha \rfloor = k\alpha - \{ k\alpha \},

where {x}\{ x \} denotes the fractional part of xx, given by {x}=x−⌊x⌋\{ x \} = x - \lfloor x \rfloor.

Hence, the sum SnS_n becomes:

Sn=(α+2α+⋯+nα)−({α}+{2α}+⋯+{nα}) S_n = (\alpha + 2\alpha + \cdots + n\alpha) - (\{ \alpha \} + \{ 2\alpha \} + \cdots + \{ n\alpha \})

Sn=n(n+1)2α−∑k=1n{kα} S_n = \frac{n(n+1)}{2}\alpha - \sum_{k=1}^{n} \{ k\alpha \}

For SnS_n to be a multiple of nn, ∑k=1n{kα}\sum_{k=1}^{n} \{ k\alpha \} must also satisfy some divisibility condition.

### Step 2: Consider Specific Values of α\alpha

- **Integer α\alpha:** If α\alpha is an integer, then each ⌊kα⌋=kα\lfloor k\alpha \rfloor = k\alpha and thus Sn=α(1+2+⋯+n)=αn(n+1)2S_n = \alpha(1 + 2 + \cdots + n) = \alpha \frac{n(n+1)}{2}, which is a multiple of nn.

- **Non-integer α\alpha:** Suppose α=m+β\alpha = m + \beta, where mm is an integer and 0<β<10 < \beta < 1. Then

⌊kα⌋=mk+⌊kβ⌋ \lfloor k\alpha \rfloor = m k + \lfloor k \beta \rfloor

For SnS_n to be a multiple of nn, the fractional parts ∑k=1n⌊kβ⌋\sum_{k=1}^{n} \lfloor k\beta \rfloor must combine to form such a multiple. However, determining this condition to hold depends intricately on β\beta.

### Step 3: Test for Simplicity with β=0\beta = 0

When β=0\beta = 0, α=2m\alpha = 2m, where 2m2m is even, the floor function simplifies without fractional interference:

Sn=(1+2+⋯+n)⋅2m=mn(n+1), S_n = (1 + 2 + \cdots + n) \cdot 2m = mn(n+1),

which is clearly a multiple of nn.

### Conclusion

From this analysis, we conclude that the condition for SnS_n to be a multiple of nn for any positive integer nn holds true for even integer values of α\alpha. Thus, the final solution is:

All even integers satisfy the condition of the problem, and no other real number α does so. \boxed{\text{All even integers satisfy the condition of the problem, and no other real number } \alpha \text{ does so.}}

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Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.