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If $x$ is an integer, find the maximum value of $$f(x)=x-\left(\lfloor r_1x\rfloor+\lfloor r_2x\rfloor+\lfloor r_3x\rfloor+...+\lfloor r_{n-2}x\rfloor+\lfloor r_{n-1}x\rfloor+\lfloor r_nx\rfloor\right)$$ Given that $\sum r_i = 1$, where $r_i \in \mathbb{Q} \space\forall i$.

So, first I applied the identity of $\lfloor r_ix\rfloor = r_ix - \{r_ix\}$ $$f(x)=x-\left(x\left(\sum r_i\right) - \left(\sum \{r_ix\}\right)\right)$$ And since $r_i$ sums to $1$, all we need to do is find the maximum value of $$f(x) =\sum \{r_ix\}$$ But $\{r_ix\} \in [0,1)$, and since we are summing from $i=1$ to $n$, and thus, there are $n$ terms in the sum, I'm tempted to say the maximum is $n-1$. A maximum value of $n$ would imply the upper bound on the fractional function is closed. Is that accurate?

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  • $\begingroup$ Can the $r_i$s be irrational? $\endgroup$
    – Erick Wong
    Commented Aug 10, 2015 at 18:29
  • $\begingroup$ @ErickWong, I forgot to stipulate that $r_i$ are rationals. Sorry, I'll fix the question! $\endgroup$
    – Ryan Tamburrino
    Commented Aug 10, 2015 at 18:31
  • $\begingroup$ Thanks, you should also add that $r_i \ge 0$ which you use later on but isn't specified by the question. Also, the question is phrased a bit oddly: $x$ is given, and you are asking for the max value of $f(x)$ which depends on $r_i$. It sounds like you are more interested in the case where the $r_i$ are given and you have the freedom to choose $x$ to maximize $f(x)$. Finally, I suspect you mean $x \in \mathbb N$ else this problem is a lot easier... $\endgroup$
    – Erick Wong
    Commented Aug 11, 2015 at 0:44
  • $\begingroup$ Yes, I believe this question was created with the intent of choosing an $x$ to maximize $f(x)$, but the problem specified only that $x \in \mathbb{Z}$. $\endgroup$
    – Ryan Tamburrino
    Commented Aug 11, 2015 at 2:37
  • $\begingroup$ Note that $f(-1) = n-1$ (assuming that $r_i > 0$). $\endgroup$
    – Erick Wong
    Commented Aug 11, 2015 at 6:47

1 Answer 1

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Trivially, $f(-1) = n-1$ provided all of the $r_i$ are non-zero. If $n=1$ this is obvious since $r_1 = 1$, and if $n>1$ then $0 < r_i < 1$ so $\lfloor -r_i \rfloor = 0$. This holds even for positive real $r_i$. When the $r_i$ are rational, then one can find a positive integer $x$ such that $f(x) = f(-1)$.

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