Find all 6-digit numbers $a_1a_2a_3a_4a_5a_6$ formed by using the digits $1,2,3,4,5,6$ once each such that the number $a_1a_2a_2\ldots a_k$ is divisible by $k$ for $1 \leq k \leq 6$.
To find all 6-digit numbers \(a_1a_2a_3a_4a_5a_6\) formed by using the digits \(1, 2, 3, 4, 5, 6\) once each such that the number \(a_1a_2a_3\ldots a_k\) is divisible by \(k\) for \(1 \leq k \leq 6\), we can follow these steps:

1. **Divisibility by 5:**
   Since \(a_1a_2a_3a_4a_5\) must be divisible by 5, \(a_5\) must be 5. Therefore, we have:
   \[
   a_5 = 5
   \]

2. **Divisibility by 3:**
   The sum of the digits \(1 + 2 + 3 + 4 + 5 + 6 = 21\) is divisible by 3. Therefore, any permutation of these digits will be divisible by 3. Hence, we need to ensure that the number \(a_1a_2a_3a_4a_5a_6\) is divisible by 6, which requires \(a_6\) to be even. Thus:
   \[
   a_6 \in \{2, 4, 6\}
   \]

3. **Divisibility by 2:**
   For \(a_1a_2\) to be divisible by 2, \(a_2\) must be even. Therefore:
   \[
   a_2 \in \{2, 4, 6\}
   \]

4. **Divisibility by 4:**
   For \(a_1a_2a_3a_4\) to be divisible by 4, \(a_4\) must be even, and the number formed by \(a_3a_4\) must be divisible by 4. Since \(a_3\) is odd (as all even numbers are used in \(a_2\), \(a_4\), and \(a_6\)), we have:
   \[
   a_4 \in \{2, 6\}
   \]

5. **Divisibility by 3 (continued):**
   For \(a_1a_2a_3\) to be divisible by 3, the sum \(a_1 + a_2 + a_3\) must be a multiple of 3. Given \(a_1\) and \(a_3\) are odd, and \(a_2\) is even, we need to check the possible values:
   \[
   a_1 + a_3 = 1 + 3 = 4 \quad \text{or} \quad 3 + 1 = 4
   \]
   Therefore, \(a_2\) must be 2 to make the sum \(a_1 + a_2 + a_3 = 6\), which is a multiple of 3:
   \[
   a_2 = 2
   \]

6. **Determine \(a_4\):**
   Using the previous steps, we have \(a_4\) as 6:
   \[
   a_4 = 6
   \]

7. **Determine \(a_6\):**
   Using the previous steps, we have \(a_6\) as 4:
   \[
   a_6 = 4
   \]

8. **Determine \(a_1\) and \(a_3\):**
   The remaining digits are 1 and 3, which can be arranged in two ways:
   \[
   a_1 = 1, a_3 = 3 \quad \text{or} \quad a_1 = 3, a_3 = 1
   \]

Thus, the two possible numbers are:
\[
123654 \quad \text{and} \quad 321654
\]

The final answer is \(\boxed{123654, 321654}\).