For the all $(m,n,k)$ positive integer triples such that $|m^k-n!| \le n$ find the maximum value of $\frac{n}{m}$

[i]Proposed by Melih Üçer[/i]
1. **Verification of the initial claim:**
   We need to verify that the triples \((m, n, k) = (1, 2, k)\) satisfy the inequality \(|m^k - n!| \le n\) for all possible values of \(k\).

   For \(m = 1\), \(n = 2\), and any \(k\):
   \[
   |1^k - 2!| = |1 - 2| = 1 \le 2
   \]
   This confirms that the inequality holds for these values.

2. **Assumption and contradiction approach:**
   Consider a triple \((m, n, k)\) that satisfies the inequality, such that \(n \geq 2m\). Suppose \(|m^k - n!| = n - l\) where \(l \in \{1, 2, \ldots, n-1\}\). There are two possible cases:

   - **Case 1: \(m^k - n! = n - l\)**
     \[
     m^k = n! + (n - l)
     \]
     This implies:
     \[
     m^k = (n - l) \left(\frac{n!}{n - l} + 1\right)
     \]
     Since \(n \geq 2m\), \(n!\) contains at least two multiples of \(m\). Thus, \(m \mid \frac{n!}{n - l}\). Suppose \(\frac{n!}{n - l} = am\).

     From the equality:
     \[
     m^k = (n - l) \left(am + 1\right)
     \]
     We obtain:
     \[
     am + 1 \mid m^k
     \]
     Take a common prime factor \(p\) of \(am + 1\) and \(m^k\). Then:
     \[
     p \mid m \Rightarrow p \mid am \Rightarrow p \mid 1
     \]
     This is a contradiction since \(p\) cannot divide 1 if \(am + 1 > 1\).

   - **Case 2: \(m^k - n! = l - n\)**
     \[
     m^k = n! - (n - l)
     \]
     This implies:
     \[
     m^k = (n - l) \left(\frac{n!}{n - l} - 1\right)
     \]
     Similarly, \(m \mid \frac{n!}{n - l}\). Suppose \(\frac{n!}{n - l} = am\).

     From the relation:
     \[
     m^k = (n - l) \left(am - 1\right)
     \]
     We deduce:
     \[
     am - 1 \mid m^k
     \]
     Hence, either \(am - 1 = 1\) or there exists a prime \(p\) which divides both \(am - 1\) and \(m^k\). The latter is rejected similarly as in the first case.

     Thus:
     \[
     am - 1 = 1 \Rightarrow am = 2 \Rightarrow n! = 2(n - l)
     \]
     This holds only for \((n, l) = (2, 1)\) or \((3, 0)\).

     However, for \(n = 3\), there are no acceptable triples \((m, n, k)\) with \(n \geq 2m\). For \(n = 2\), the triples are all of the form \((m, n, k) = (1, 2, k)\) and only them.

3. **Conclusion:**
   Thus, the maximum value of the ratio \(\frac{n}{m}\) is indeed equal to 2 and is achieved only for \(m = 1, n = 2\).

The final answer is \(\boxed{2}\)