Prove that for any $x \in \mathbb N$ such that $xProve that every positive integer $x$ with $x<n!$ is the sum of at most $n$ distinct divisors of $n!$.
 A: People do not seem to be going along with my comment. So this is CW, and directly from the answer by Sanchez.
For $n=2,$ we need only 1 divisor of $2!,$ as $1=1.$
For $n=3,$ we need only 2 divisors of $3!=6,$ as $1=1, 2=2,3=3,4=3+1,5=3+2.$
Induction hypothesis: for some $n \geq 2,$ we need at most $(n-1)$ distinct divisors of $n!$ to write any $1 \leq x < n!$ as a sum.
Induction step (Sanchez, above). Let $N = n+1.$ Let $1 \leq x < N! = (n+1)!$ Write
$$  x = N q + r, \; \; \mbox{with} \; \; 0 \leq r < N.   $$
Because $q < (N-1)! = n!,$ we need at most $(n-1) = (N-2)$ divisors of $n!$ to write $q$ as a sum. So
$$ q = \sum_{i=1}^{n-1} d_i,   $$
where each $d_i | n!$ Therefore each $Nd_i | N!$
At this stage, we have at most $N-2$ divisors of $N!$ What about $r?$ Well, $r < N,$ so it is automatically a factor of $N!$ So we have finished the decomposition as a sum with at most $(N-1)$ divisors of $N!,$ where $N=n+1.$
CONCLUSION: For all $N \geq 2,$ every integer $1 \leq x < N!$ can be written as the sum of (at most) $N-1$ distinct divisors of $N!$
SUGGESTION: try it for $N=4, \; \; N! = 24.$
NEVER MIND, do it myself. Aliquot divisors 1,2,3,4,6,8,12. 
$$1=1,2=2,3=3,4=4,5=4+1,6=4+2,7=4+3, 8=8,9=8+1,10=8+2,  $$
$$11=8+3, 12= 12, 13 = 12+1, 14 = 12+2, 15 = 12+3, 16 = 12+4,$$
$$17=12+4+1,   18=12+6,19=12+4+3,20=12+8,$$
$$21=12+8+1,22=12+8+2,23 = 12+8+3.   $$
A: Hint: Note that $x = m (n-1)! + r$ where $0 \le m < n$ and $0 \le r < (n-1)!$.  Use induction.
EDIT: Oops, this is wrong: as Steven Stadnicki noted, $m (n-1)!$ doesn't necessarily divide $n!$.
A: Let $x = nq+r$, with $0 \leq r < n$. Note that $x < n!$ implies that $q < (n-1)!$. Now use induction on $q$.
A: Not quite there, but a start:  As suggested in Wikipedia on practical numbers we will use the greedy algorithm.  First pull out $n!/2$ if that is possible, then $n!/3$, then $n!/4$ and so on, stopping when the remainder is less than or equal to $n$ and skipping denominators that don't divide $n!$.  If $n$ and $x$ are very large, the denominators we use will follow Sylvester's sequence:  $2, 3, 7, 43, 1807, 3263443, 10650056950807,\ldots$ which is given by $a(0)=2, a(n+1)=a(n)^2-a(n)+1$.  To use induction, we need to find a sequence of $m$ denominators that reduce $n!-1$ to something less than $n$.  For $n$ in the range $5-6$ we can use $2,3,8,30$.  For $7$ we can use $2,3,7,45$, for $8-10$ we can use $2,3,7,45,640$.  Then $44$ becomes available at $11$.  It "obviously" works, but I can't prove it.
A: A natural number $m$ is called practical if all smaller natural numbers can be represented as sum of distinct divisors of $m$.
The problem asks to establish that factorial numbers are practical. The wikipedia article on practical numbers even gives an algorithm, implemented in Mathematica:
dec2[0, n_] := {};
dec2[1, n_] := {1};
dec2[x_, n_] := Module[{fcts, pa, q, r, quo},
  fcts = Last[FactorInteger[n]]; pa = Power @@ fcts;
  q = Min[Quotient[x, pa], DivisorSigma[1, quo = Quotient[n, pa]]];
  Join[dec2[x - q pa, Quotient[n, First[fcts]] ], dec2[q, quo] pa]
  ]

dec[x_, n_] := Block[{$RecursionLimit = Infinity}, dec2[x, n!]]

Example:
In[32]:= dec[17, 4]

Out[32]= {2, 3, 12}

In[33]:= dec[137, 6]

Out[33]= {2, 45, 90}

It remains to be proven that the decomposition length of $x < n!$ will be less of equal than $n$.
