# Prove that for any $x \in \mathbb N$ such that $x<n!$ is the sum of at most $n$ distinct divisors of $n!$

Prove that for any $x \in \mathbb N$ such that $x<n!$ is the sum of at most $n$ distinct divisors of $n!$.

-
What have you tried? –  Matthew Conroy Aug 2 '12 at 18:07
If only Goldbach conjecture was true. –  Jayesh Badwaik Feb 20 '13 at 2:55

Let $x = nq+r$, with $0 \leq r < n$. Note that $x < n!$ implies that $q < (n-1)!$. Now use induction on $q$.

-
Works for me. Note that, as the initial case is 2! rather than 1!, this proves that $(n-1)$ divisors suffice. –  Will Jagy Feb 20 '13 at 3:34
Great, may I ask if you had seen a similar problem before or how you figured it out. Thanks! –  dREaM Feb 20 '13 at 3:56
@Khromonkey, I haven't seen this before. Induction seems to be a natural idea though. I guess that Robert Israel's attempt helped inspire this solution too. –  user27126 Feb 20 '13 at 4:04
No, this doesn't work, either. Let $n=5, x=119$, then $q=23$ and $115$ doesn't divide $120$ –  Ross Millikan Feb 20 '13 at 13:42
@Ross Millikan, $q = 23 < 4!$, and you want to cut 23 up into sum of divisors of 4! first. For example, 23 = 12 + 8 + 3, and 119 = 60 + 40 + 15 + 4. –  user27126 Feb 20 '13 at 16:33

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. $$1=1,2=2,3=3,4=4,5=4+1,6=4+2,7=4+3, 8=8,9=8+1,10=8+2,$$ $$11=8+3, 12= 12, 13 = 12+1, 14 = 12+2, 15 = 12+3, 16 = 12+4,$$ $$17=12+4+1, 18=12+6,19=12+4+3,20=12+8,$$ $$21=12+8+1,22=12+8+2,23 = 12+8+3.$$

-

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 $n!$.

-
$m(n-1)!$ isn't necessarily a divisor of $n!$, is it? –  Steven Stadnicki Aug 2 '12 at 19:04
@RobertIsrael Then what is the correct answer? –  dREaM Sep 22 '12 at 16:55
I don't know. I believe it is true, and probably not too hard to prove, that for any $x \in \{1,\ldots,n!\}$ there is always a divisor $y$ of $n!$ with $x/2 \le y \le x$. Then we can take a greedy approach: let $y_1$ be the greatest divisor of $n!$ less than $x$, and replace $x$ by $x - y_1$. This will result in writing $x$ as a sum of distinct divisors of $n!$, but it's not clear to me that there will be at most $n$ of them: the easy bound would be about $\log_2(n!) \approx n \log_2(n)$. –  Robert Israel Sep 23 '12 at 18:07
Can you edit or delete your answer please?? If it is wrong why is it here?? –  dREaM Feb 19 '13 at 22:55
@Khromonkey, it is probably here because it is a reasonable direction to take, yet involves an error that has been identified. As such, it is educational for others. Meanwhile, noting that you say you are a high-school student, I cannot see how this is an ordinary homework problem, so I am asking why you think your assertion is true and where you got the problem. –  Will Jagy Feb 19 '13 at 23:35

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 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\$.

-
That is the question. Can you make this answer answer it? –  dREaM Feb 19 '13 at 22:54