Find the largest integer $n \ge 3$ for which there is a $n$-digit number $\overline{a_1a_2a_3...a_n}$ with non-zero digits
$a_1, a_2$ and $a_n$, which is divisible by $\overline{a_2a_3...a_n}$.
1. **Understanding the problem:**
   We need to find the largest integer \( n \ge 3 \) such that there exists an \( n \)-digit number \(\overline{a_1a_2a_3\ldots a_n}\) with non-zero digits \(a_1, a_2, \ldots, a_n\), which is divisible by \(\overline{a_2a_3\ldots a_n}\).

2. **Setting up the problem:**
   Let \(\overline{a_2a_3\ldots a_n} = c\) and \(a_1 = a\). Then, the \( n \)-digit number can be written as:
   \[
   \overline{a_1a_2a_3\ldots a_n} = 10^{n-1}a + c
   \]
   We need \(c\) to divide \(10^{n-1}a + c\), which implies:
   \[
   c \mid 10^{n-1}a
   \]
   Since \(c\) is a number formed by the digits \(a_2, a_3, \ldots, a_n\), it cannot be divisible by 10 (as \(a_n \neq 0\)).

3. **Analyzing the divisibility:**
   We have three cases to consider:
   - \(2 \nmid c\) and \(5 \nmid c\)
   - \(2 \mid c\) and \(5 \nmid c\)
   - \(5 \mid c\) and \(2 \nmid c\)

4. **Case 1: \(2 \nmid c\) and \(5 \nmid c\):**
   If \(c\) is not divisible by 2 or 5, then \(c\) must divide \(a\). Since \(a\) is a single digit (1 to 9), \(c\) must be a single-digit number. This implies \(n = 2\), which contradicts \(n \ge 3\).

5. **Case 2: \(2 \mid c\) and \(5 \nmid c\):**
   If \(c\) is divisible by 2 but not by 5, then \(c\) must divide \(2^{n-1}a\). We need:
   \[
   10^{n-2} < c \leq 2^{n-1}a \leq 9 \cdot 2^{n-1}
   \]
   This implies:
   \[
   5^{n-2} < 18 \implies n \leq 3
   \]
   This does not provide a solution for \(n \ge 3\).

6. **Case 3: \(5 \mid c\) and \(2 \nmid c\):**
   If \(c\) is divisible by 5 but not by 2, then \(c\) must divide \(5^{n-1}a\). We need:
   \[
   10^{n-2} < c \leq 9 \cdot 5^{n-1}
   \]
   This implies:
   \[
   2^{n-2} < 45 \implies n \leq 7
   \]

7. **Checking \(n = 7\):**
   For \(n = 7\), we need \(c\) to divide \(5^6a\). When \(a = 9\), \(c\) must divide \(140625\). However, \(c > 10^5\) and \(c = 140625\) contains a zero, which is not allowed. Similarly, for \(a = 8\) and \(a = 7\), \(c = 125000\) and \(c = 109375\) respectively, both contain zeros. For \(a < 7\), \(c\) cannot be greater than \(10^5\).

8. **Conclusion:**
   Since \(n = 7\) does not provide a valid solution, we check \(n = 6\). For \(n = 6\), we find:
   \[
   928125 = 33 \cdot 28125
   \]
   Here, \(28125\) is a valid \(c\) and does not contain any zeros.

The final answer is \( \boxed{ 6 } \).