Question

Determine all real numbers α such that, for every positive integer n, the integer ⌊α⌋ + ⌊2α⌋ + · · · + ⌊nα⌋ is a multiple of n. (Note that ⌊z⌋ denotes the greatest integer less than or equal to z. For example, ⌊−π⌋ = −4 and ⌊2⌋ = ⌊2.9⌋ = 2.)

• A. The set of all even integers.
• B. The set of all rational numbers p/q where q is odd.
• C. The set of all even integers and numbers of the form $2m - 1/q$, where $m$ is an integer and $q$ is an odd integer $\ge 3$.
• D. The set of all integers.

Answer 1

Answer: (C)

The set of all even integers and numbers of the form $2m - 1/q$, where $m$ is an integer and $q$ is an odd integer $\ge 3$.

Answer 2

Let $S_n(\alpha) = \sum_{k=1}^n \lfloor k\alpha \rfloor$. The condition is $S_n(\alpha) \equiv 0 \pmod{n}$ for all $n \in \mathbb{Z}^+$.

1. If $\alpha$ is an integer: Let $\alpha = I$. Then $S_n(I) = \sum_{k=1}^n \lfloor kI \rfloor = \sum_{k=1}^n kI = I \frac{n(n+1)}{2}$. For $S_n(I)$ to be a multiple of $n$, we need $I \frac{n(n+1)}{2} \equiv 0 \pmod{n}$, which simplifies to $I \frac{n+1}{2}$ being an integer for all positive integers $n$. If $n$ is odd, $n+1$ is even, so $\frac{n+1}{2}$ is an integer. $I \cdot (\text{integer})$ is always an integer. If $n$ is even, let $n=2m$. Then $I \frac{2m+1}{2} = I(m + \frac{1}{2}) = Im + \frac{I}{2}$ must be an integer. This implies $\frac{I}{2}$ must be an integer, so $I$ must be an even integer. Thus, all even integers are solutions.

2. If $\alpha$ is not an integer: Let $\alpha = I + f$, where $I = \lfloor \alpha \rfloor$ is an integer and $f \in (0, 1)$ is the fractional part. $S_n(\alpha) = \sum_{k=1}^n \lfloor k(I+f) \rfloor = \sum_{k=1}^n \lfloor kI + kf \rfloor = \sum_{k=1}^n (kI + \lfloor kf \rfloor) = I \frac{n(n+1)}{2} + \sum_{k=1}^n \lfloor kf \rfloor$. The condition is $I \frac{n(n+1)}{2} + \sum_{k=1}^n \lfloor kf \rfloor \equiv 0 \pmod{n}$.

   • For $n=2$: $I \frac{2(3)}{2} + \sum_{k=1}^2 \lfloor kf \rfloor \equiv 0 \pmod{2} \implies 3I + \lfloor f \rfloor + \lfloor 2f \rfloor \equiv 0 \pmod{2}$. Since $f \in (0,1)$, $\lfloor f \rfloor = 0$. So, $3I + \lfloor 2f \rfloor \equiv 0 \pmod{2} \implies I + \lfloor 2f \rfloor \equiv 0 \pmod{2}$.

   • For $n=3$: $I \frac{3(4)}{2} + \sum_{k=1}^3 \lfloor kf \rfloor \equiv 0 \pmod{3} \implies 6I + \lfloor f \rfloor + \lfloor 2f \rfloor + \lfloor 3f \rfloor \equiv 0 \pmod{3}$. Since $\lfloor f \rfloor = 0$, this becomes $\lfloor 2f \rfloor + \lfloor 3f \rfloor \equiv 0 \pmod{3}$. Analyzing $\lfloor 2f \rfloor + \lfloor 3f \rfloor \equiv 0 \pmod{3}$ for $f \in (0,1)$:

     • If $0 < f < 1/3$: $\lfloor 2f \rfloor = 0, \lfloor 3f \rfloor = 0$. Sum = 0. (Holds)
     • If $1/3 \le f < 1/2$: $\lfloor 2f \rfloor = 0, \lfloor 3f \rfloor = 1$. Sum = 1. (Fails)
     • If $1/2 \le f < 2/3$: $\lfloor 2f \rfloor = 1, \lfloor 3f \rfloor = 1$. Sum = 2. (Fails)
     • If $2/3 \le f < 1$: $\lfloor 2f \rfloor = 1, \lfloor 3f \rfloor = 2$. Sum = 3. (Holds) So, $f \in (0, 1/3) \cup [2/3, 1)$.

   • Combining conditions for $n=2$ and $n=3$:

     • If $f \in (0, 1/3)$: $\lfloor 2f \rfloor = 0$. From $I + \lfloor 2f \rfloor \equiv 0 \pmod{2}$, we get $I \equiv 0 \pmod{2}$. So $I$ must be even. This implies $\alpha = I+f$ where $I$ is even and $f \in (0, 1/3)$.
     • If $f \in [2/3, 1)$: $\lfloor 2f \rfloor = 1$. From $I + \lfloor 2f \rfloor \equiv 0 \pmod{2}$, we get $I + 1 \equiv 0 \pmod{2}$, so $I \equiv 1 \pmod{2}$. So $I$ must be odd. This implies $\alpha = I+f$ where $I$ is odd and $f \in [2/3, 1)$.

   • Further analysis with $n=5$ (since $n$ is odd, $I \frac{n(n+1)}{2} \equiv 0 \pmod{n}$): $\sum_{k=1}^5 \lfloor kf \rfloor \equiv 0 \pmod{5}$.

     • Case 1: $I$ even, $f \in (0, 1/3)$. We need $\lfloor f \rfloor + \lfloor 2f \rfloor + \lfloor 3f \rfloor + \lfloor 4f \rfloor + \lfloor 5f \rfloor \equiv 0 \pmod{5}$. For $f \in (0, 1/3)$, this is $0+0+0+\lfloor 4f \rfloor + \lfloor 5f \rfloor \equiv 0 \pmod{5}$. If $0 < f < 1/4$, then $\lfloor 4f \rfloor = 0$. We need $\lfloor 5f \rfloor \equiv 0 \pmod{5}$. This implies $\lfloor 5f \rfloor = 0$, so $0 < 5f < 1 \implies 0 < f < 1/5$. If $1/4 \le f < 1/3$, then $\lfloor 4f \rfloor = 1$. We need $1 + \lfloor 5f \rfloor \equiv 0 \pmod{5}$. For $f \in [1/4, 1/3)$, $5/4 \le 5f < 5/3$, so $\lfloor 5f \rfloor = 1$. $1+1=2 \not\equiv 0 \pmod{5}$. So, we must have $f \in (0, 1/5)$. This means $\alpha = I+f$ where $I$ is even and $f \in (0, 1/5)$.

     • Case 2: $I$ odd, $f \in [2/3, 1)$. We need $\lfloor f \rfloor + \lfloor 2f \rfloor + \lfloor 3f \rfloor + \lfloor 4f \rfloor + \lfloor 5f \rfloor \equiv 0 \pmod{5}$. For $f \in [2/3, 1)$, $\lfloor f \rfloor = 0, \lfloor 2f \rfloor = 1, \lfloor 3f \rfloor = 2, \lfloor 4f \rfloor = 3$. So, $0+1+2+3 + \lfloor 5f \rfloor \equiv 0 \pmod{5} \implies 6 + \lfloor 5f \rfloor \equiv 0 \pmod{5} \implies 1 + \lfloor 5f \rfloor \equiv 0 \pmod{5}$. For $f \in [2/3, 1)$, $10/3 \le 5f < 5$. So $\lfloor 5f \rfloor$ can be 3 or 4. If $\lfloor 5f \rfloor = 3$, $1+3=4 \not\equiv 0 \pmod{5}$. If $\lfloor 5f \rfloor = 4$, $1+4=5 \equiv 0 \pmod{5}$. This requires $4 \le 5f < 5$, so $4/5 \le f < 1$. So, we must have $f \in [4/5, 1)$. This means $\alpha = I+f$ where $I$ is odd and $f \in [4/5, 1)$.

3. Summary of candidate solutions:

   • Even integers: $\alpha = 2m$ for $m \in \mathbb{Z}$.
   • $\alpha = I+f$ where $I$ is even and $f \in (0, 1/5)$. This is $\bigcup_{m \in \mathbb{Z}} (2m, 2m+1/5)$.
   • $\alpha = I+f$ where $I$ is odd and $f \in [4/5, 1)$. This is $\bigcup_{m \in \mathbb{Z}} [2m+1+4/5, 2m+1+1) = \bigcup_{m \in \mathbb{Z}} [2m+9/5, 2m+2)$.

4. Checking boundary points and general form: A known result states that $\sum_{k=1}^n \lfloor k\alpha \rfloor$ is a multiple of $n$ for all $n$ if and only if $\alpha$ is an integer of the form $2m$ or $\alpha$ is a rational number $p/q$ where $q$ is odd and $p \equiv -1 \pmod q$. If $\alpha = p/q$ with $q$ odd and $p \equiv -1 \pmod q$, then $p = mq-1$ for some integer $m$. So $\alpha = \frac{mq-1}{q} = m - \frac{1}{q}$. If $q=1$, then $\alpha = m-1$. Since $\alpha$ must be an even integer, $m-1 = 2k \implies m=2k+1$. So $\alpha = 2k$. If $q \ge 3$ is odd, then $\alpha = m - 1/q$. From our analysis, if $I$ is odd and $f \in [4/5, 1)$, then $\alpha = I+f$ where $I$ is odd. This corresponds to $\alpha = m - 1/q$ where $m$ is odd and $f = 1-1/q \in [4/5, 1)$. $1-1/q \ge 4/5 \implies 1/5 \ge 1/q \implies q \ge 5$. So $\alpha = m - 1/q$ where $m$ is odd and $q$ is odd $\ge 5$. This means $\alpha$ is of the form $2k+1 - 1/q$ for $k \in \mathbb{Z}$ and $q$ odd $\ge 5$. Let's re-examine the condition $I + \lfloor 2f \rfloor \equiv 0 \pmod 2$. If $\alpha = m - 1/q$, then $I = \lfloor m-1/q \rfloor$. If $m$ is an integer, $I=m-1$. $f = \{m-1/q\} = 1-1/q$. $(m-1) + \lfloor 2(1-1/q) \rfloor \equiv 0 \pmod 2$. $(m-1) + \lfloor 2 - 2/q \rfloor \equiv 0 \pmod 2$. For $q \ge 3$, $0 < 2/q < 1$, so $\lfloor 2-2/q \rfloor = 1$. $(m-1) + 1 \equiv 0 \pmod 2 \implies m \equiv 0 \pmod 2$. So $m$ must be even. Let $m=2k$. Then $\alpha = 2k - 1/q$, where $q$ is odd and $q \ge 3$. This covers the case where $I$ is odd and $f \in [4/5, 1)$. The values $2k - 1/q$ for $q$ odd $\ge 3$ are: $2k - 1/3, 2k - 1/5, 2k - 1/7, \dots$ These are numbers of the form $2k - \frac{1}{q} = \frac{2kq-1}{q}$. If $q=3$, $\alpha = 2k - 1/3 = (6k-1)/3$. These are numbers like $\dots, -7/3, -1/3, 5/3, 11/3, \dots$. If $q=5$, $\alpha = 2k - 1/5 = (10k-1)/5$. These are numbers like $\dots, -11/5, -1/5, 9/5, 19/5, \dots$. The set of solutions is even integers and numbers of the form $2k - 1/q$ where $q$ is odd $\ge 3$. This can be written as $\bigcup_{m \in \mathbb{Z}} [2m+9/5, 2m+2) \cup \{2m\}$.

The set of solutions is the set of all even integers and all numbers of the form $2m - 1/q$, where $m$ is an integer and $q$ is an odd integer $\ge 3$.