Integers inequality I have found this problem in another site: I have tried to solve it but without success !.


*

*A large industrial site has 5000 workers. For each of them, the sum of his direct superiors and subordinates is $7$. 

*Every day there are several work orders, for which the pattern of issuance, communication and execution is as follows, for every day of the week:


*

*Every Monday, each worker issues a work order and distributes copies of it to his direct subordinates ( if, of course, he has any - otherwise he executes them himself ).

*Each Tuesday, all workers that received work orders on Monday, distribute them to their direct subordinates, if any; otherwise they execute them themselves.

*Each Wednesday, procedure number 2. is repeated: Any workers that received work orders on Tuesday, distribute them to their subordinates, otherwise they execute them themselves. Same also on Thursday and Friday.
Finally, on Friday there are no more work orders for distribution to any subordinates ( that is, any remaining work orders will be executed by the workers themselves ).  



What is the MINIMUM number of workers that do NOT have direct superiors ?. Note that one worker may have shared subordinates with some other worker(s).
If we name the minimum number of employees without direct superiors $x$, and then they distribute their work orders to $y$, we must have: $x \leq y \leq 7x$.
Then they distribute each subsequent day to $z$, $v$ and $w$. The total of all must be $5000$.
We must somehow work out the inequalities and keep only the integer solutions.
I think the inequalities are:
\begin{align}
&x \leq y \leq 7x\,,\quad
7y \geq x + z\,,\quad
7z \geq y + v\,,\quad
7v \geq z + w
\\[2mm]
&
7w = v\qquad \left(~\mbox{since on Friday we do not have any work orders to distribute}~\right).
\\[2mm]
&\mbox{Also,}\quad x + y + z + v + w = 5000.
\end{align}
Does this make any sense ?. I don't know how to continue. 
 A: So the problem tells that the hierachical chain has $5$ levels. 
Suppose there is just $x=1$ top-director;       
then he must have $y=7$ 2nd level directors;    
suppose each of these to have a separated subordinate chain, 
so each can have max $6$ subordinates, which makes $z=7*6$;   
same for the following level, to give $v=7*6*6$;    
finally suppose that each $v$ also has a separated subordinate chain, each must be of $6$ dependents, so
 $w=7*6*6*6$.
But, with this scheme, each employee at the lowest level  refers to only $1$ direct boss, while (not having any lower) it should be seven.
(this is what the problem literally establishes: exactly $7$ between upper and lower).
So forget for a moment the scheme above and suppose to have just $m$ employees at $v$,
those in the yellow box, in the following scheme.

To each of these we shall assign $6$ different employees at level $w$ :
 $(w_{1,1}, \cdots, w_{1,6}),\cdots, (w_{m,1}, \cdots, w_{m,6})$.
This in order to maximize the global number, so the $v$'s are still taken to have
 one "upper" bond (at the moment left open) and six lower ones.
And to each $w_{j,k}$ we shall assign $6$ additional bosses.
The previous $v$ are already "full", so we shall introduce other bosses at $v$.
These also shall have bounds $1-6$.
The two requirements ($6$ additional bosses / $6$ lower employees) can be 
matched if we assign a $w$ sextuple to a sextuple of additional bosses.
Therefore we need $6m$ additional bosses, i.e. a total of $7m$ at $v$
vs. a total of $6m$ at $w$.
Intuitively it looks that the scheme ensures the max possible value for $v$ and $w$,
but I am unable to prove that rigorously.
Now let's recall the previous scheme, where
starting with $x=1$ we got $v=7*6*6$. This means that we can concile the two schemes taking $m=36$ and $w=6*m=216$.
And $x+y+z+v+w = 1+7+42+252+216 = 518$.  
Thus
$$
\left\lceil {{{5000} \over {518}}} \right\rceil  = 10 \le x
$$
