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Let $n\geq 2$ and $x_1,\ldots,x_n > 0$ be such that $x_1+\cdots+x_n =1$. Is it true that there must exist a positive integer $k$ such that $$\{x_1k\}+\cdots+\{x_nk\} = n-1?$$

This looks closely related to the density of the fractional part. Note that the quantity $\{x_1k\}+\cdots+\{x_nk\}$ is always an integer, since it equals $k-\lfloor x_1k\rfloor - \dots - \lfloor x_nk\rfloor$. Also, as each term is strictly less than one, $n-1$ is the highest value the sum can take.

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If the $x_i$ are rational, take the lcm of the denominators and decrease it by $1$.

If they are not necessarily rational, act similarly: using, e.g., Kronecker's theorem, take a $k$ such that all the $kx_i$ are sufficiently close to integers, and decrease that $k$ by $1$.

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  • $\begingroup$ I've put a reference, although it is an overkill: one may merely use the pigeonhole principle. $\endgroup$
    – Ilya Bogdanov
    Commented May 10, 2020 at 14:50
  • $\begingroup$ ... another way of saying it is that the curve $t\mapsto (tx_1,\ldots ,tx_n)$ is dense in the torus $\mathbb{R}^n/\mathbb{Z}^n$, thus becomes as close as you want to $0$ for some $t>0$. $\endgroup$
    – abx
    Commented May 10, 2020 at 14:59
  • $\begingroup$ @abx: this is incorrect as soon as $1,x_1,x_2,\dots,x_n$ are linearly dependent over $\mathbb Q$. That's why Kronecker's theorem at my reference looks a bit... involved. In our case, they are dependent, as the $x_i$ sum up to $1$! $\endgroup$
    – Ilya Bogdanov
    Commented May 10, 2020 at 19:14
  • $\begingroup$ Oh, right of course. Still the curve is dense in some nontrivial subtorus (if at least one of the $x_i$ is irrational). But that makes the proof a bit complicated. $\endgroup$
    – abx
    Commented May 11, 2020 at 4:37
  • $\begingroup$ @IlyaBogdanov what does "as soon as" mean? $\endgroup$
    – mathworker21
    Commented May 16, 2020 at 18:24

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