Simultaneous equations with floor functions

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Joined: Jun 5, 2008

by cip1703, Jan 20, 2018, 10:49 AM

==Problem==
Let $\lfloor x \rfloor$ be the greatest integer less than or equal to $x$ and let $\{x\}=x-\lfloor x \rfloor$ . Solve

$\left\{\begin{array}{cc} x+\lfloor y \rfloor + \{z\}=a & (1)\\\\ \{x\}+ y+\lfloor z \rfloor=b & (2)\\\\ \lfloor x \rfloor+\{y\}+z=c & (3)\end{array}\right\|$

where $a$ , $b$ , $c$ are given real numbers.

==Solution==
.........ANSWER: $x=\lfloor \frac{a-b+c}{2} \rfloor+ \{\frac{a+b-c}{2}\}$ , $y=\lfloor \frac{a+b-c}{2} \rfloor+ \{\frac{-a+b+c}{2}\}$ , $z=\lfloor \frac{-a+b+c}{2}\rfloor+ \{\frac{a-b+c}{2}\}.$

...........For any $t\in\mathbb R$ we have
$t=\lfloor t \rfloor + \{t\}$ (4)
and conversely, if $t=k+\alpha, t\in\mathbb Z, \alpha \in [0;1)$ it follow that $\lfloor t \rfloor =k$ and $\{t\}=\alpha.$
Adding relations (1),(2),(3) we get $x+\lfloor y \rfloor + \{z\}+\{x\}+ y+\lfloor z \rfloor+\lfloor x \rfloor+\{y\}+z=a+b+c$
$\iff x+y+z=\frac{a+b+c}{2}$ (5).
Writing (5) as $x+\lfloor y \rfloor +\{z\}+\lfloor z \rfloor+\{y\}=\frac{a+b+c}{2}$ and taking account of (1) we get

$\lfloor z \rfloor+\{y\}=\frac{a+b+c}{2}-a=\frac{-a+b+c}{2}$ , and so $\lfloor z \rfloor=\lfloor\frac{-a+b+c}{2}\rfloor$ , and $\{y\}=\{\frac{-a+b+c}{2}\}$ .
In the same way we obtain $\lfloor x \rfloor+\{z\}=\frac{a-b+c}{2}$ , $\lfloor y \rfloor+\{z\}=\frac{a+b-c}{2}$
Taking account of (1) we have $x=\lfloor x \rfloor +\{x\}=...$ and we get the answer.

See also https://artofproblemsolving.com/texer/boixabdc

This post has been edited 12 times. Last edited by cip1703, Jan 20, 2018, 4:17 PM
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CIP1703

Simultaneous equations with floor functions

by cip1703, Jan 20, 2018, 10:49 AM

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