Matematické Fórum

stuart clark · 20. 09. 2012 18:50

Calculate value of $x$ in $\lfloor x \rfloor +\lfloor \frac{x}{2} \rfloor+\lfloor \frac{x}{3} \rfloor+\lfloor \frac{x}{4} \rfloor+\lfloor \frac{x}{5} \rfloor+\lfloor \frac{x}{6} \rfloor = 2011$

where $\lfloor x \rfloor =$ floor function.

BakyX · 20. 09. 2012 18:55

$x \in \mathbb{R}$ ?

vanok · 23. 09. 2012 16:28 — Editoval vanok (27. 09. 2012 09:20)

Hi re]p303725|BakyX[/re],

Yes, I believe that $x \in \mathbb{R}$ .

For the solution, it is sufficient to use the following inequality:
$\lfloor \frac x i \rfloor \leq \frac x i <\lfloor \frac x i \rfloor +1$ for $i \in \{1; 2; ... ; 6\}$
So, if a such x exists $2011 \leq x\cdot 2.45 <2011+6=2017$ ( What is a necessary condition, considering size of rests of $\frac x i$ )

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Finally,it is a question of finding x suitable such as
$\frac {2011 }{2.45} \leq x < \frac {2017 }{2,45}$

We notice that for x = 822 (candidate for solution), we have
$\lfloor x \rfloor +\lfloor \frac x2 \rfloor +\lfloor \frac x3 \rfloor+\lfloor \frac x4 \rfloor +\lfloor \frac x5 \rfloor +\lfloor \frac x 6 \rfloor=\lfloor 822 \rfloor +\lfloor \frac {822}2 \rfloor +\lfloor \frac {822}3 \rfloor+\lfloor \frac {822}4 \rfloor +\lfloor \frac {822}5 \rfloor +\lfloor \frac {822}6 \rfloor=$
$822+411+274+205+164+137=2013$

What shows: a posible solution x is < 822

We notice:
that the function $x\rightarrow \lfloor x \rfloor +\lfloor \frac x2 \rfloor +\lfloor \frac x3 \rfloor+\lfloor \frac x4 \rfloor +\lfloor \frac x5 \rfloor +\lfloor \frac x 6 \rfloor$ is increasing.
And
$\frac {822}2 =411 =\lfloor \frac {822}2 \rfloor$

$\frac {822}3 =274 =\lfloor \frac {822}3 \rfloor$

$\frac {822}6 =137 =\lfloor \frac {822}6 \rfloor$

So for $x=822-\varepsilon$ where $\varepsilon$ is > 0, rather small, we have
$\lfloor x \rfloor +\lfloor \frac x2 \rfloor +\lfloor \frac x3 \rfloor+\lfloor \frac x4 \rfloor +\lfloor \frac x5 \rfloor +\lfloor \frac x 6 \rfloor= 2009$
Because
$\lfloor 822-\varepsilon \rfloor=821$

$\lfloor \frac {822-\varepsilon}2 \rfloor=410$

$\lfloor \frac {822-\varepsilon}3 \rfloor=273$

$\lfloor \frac {822-\varepsilon}6 \rfloor=136$

What means that in $822$ a jump of size 4 makes.

Conclusion
the problem gives does not admit solution

Edit: Proof completed thanks to a colleague's (Honz ) constructive remark (thank you Honz )

stuart clark · 24. 09. 2012 04:23

Thanks vanok I am getting $x = 820\;,822$

vanok · 24. 09. 2012 13:53

Hi ↑ stuart clark:

I edited my explicit solution here ↑ vanok:

Matematické Fórum

#1 20. 09. 2012 18:50

floor equation

#2 20. 09. 2012 18:55

Re: floor equation

#3 23. 09. 2012 16:28 — Editoval vanok (27. 09. 2012 09:20)

Re: floor equation

#4 24. 09. 2012 04:23

Re: floor equation

#5 24. 09. 2012 13:53

Re: floor equation

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