Compute the number of integers $1 \leq n \leq 1024$ such that the sequence $\lceil n \rceil$, $\lceil n/2 \rceil$, $\lceil n/4 \rceil$, $\lceil n/8 \rceil$, $\ldots$ does not contain any multiple of $5$. 

[i]Proposed by Sean Li[/i]
To solve the problem, we need to count the number of integers \(1 \leq n \leq 1024\) such that the sequence \(\lceil n \rceil, \lceil n/2 \rceil, \lceil n/4 \rceil, \lceil n/8 \rceil, \ldots\) does not contain any multiple of 5. We will use complementary counting to find the solution.

1. **Complementary Counting**:
   We will count the number of integers \(n\) such that the sequence contains at least one multiple of 5, and then subtract this count from 1024.

2. **Form of \(n\)**:
   For \(\left \lceil \frac{n}{2^p} \right \rceil\) to be a multiple of 5, \(n\) must be of the form:
   \[
   n = k \cdot (5 \cdot 2^p) - r \quad \text{where} \quad 0 \leq r < 2^p
   \]
   This motivates us to consider all \(n\) modulo \(5, 5 \cdot 2, 5 \cdot 2^2, \ldots, 5 \cdot 2^7\) since for \(2^8\) and above there are no possible \(1 \leq n \leq 1024\).

3. **Casework on \(p\)**:
   We will consider cases based on the smallest \(p\) such that \(5 \mid \left \lceil \frac{n}{2^p} \right \rceil\) but \(5 \nmid \left \lceil \frac{n}{2^q} \right \rceil\) for \(q < p\). Denote such value as \(p_0\).

   - **Case 1: \(p_0 = 0\)**:
     \[
     n \equiv 0 \pmod{5}
     \]
     There are \(\boxed{204}\) possible values.

   - **Case 2: \(p_0 = 1\)**:
     \[
     n \equiv -1 \pmod{10}
     \]
     There are \(\boxed{102}\) possible values.

   - **Case 3: \(p_0 = 2\)**:
     \[
     n \equiv -2, -3 \pmod{20}
     \]
     There are \(\boxed{102}\) possible values.

   - **Case 4: \(p_0 = 3\)**:
     \[
     n \equiv -4, -6, -7 \pmod{40}
     \]
     There are \(\boxed{75}\) possible values.

   - **Case 5: \(p_0 = 4\)**:
     \[
     n \equiv -8, -9, -12, -13, -14 \pmod{80}
     \]
     There are \(\boxed{60}\) possible values.

   - **Case 6: \(p_0 = 5\)**:
     \[
     n \equiv -16, -17, -18, -19, -24, -26, -27, -28, -29 \pmod{160}
     \]
     There are \(\boxed{54}\) possible values.

   - **Case 7: \(p_0 = 6\)**:
     \[
     n \equiv -32, -33, -34, -36, -37, -38, -39, -48, -49, -52, -53, -54, -56, -57, -58, -59 \pmod{320}
     \]
     There are \(\boxed{48}\) possible values.

   - **Case 8: \(p_0 = 7\)**:
     Using the method mentioned in the remark to find the appropriate residues, we find that there are 100 bad residues, so there are:
     \[
     128 - 100 = 28
     \]
     possible values for \(n\). Thus, there are \(\boxed{28}\) values for \(n\).

4. **Summing Up**:
   Summing the values from all cases, we get:
   \[
   204 + 102 + 102 + 75 + 60 + 54 + 48 + 28 = 673
   \]
   Therefore, the number of integers \(n\) such that the sequence does not contain any multiple of 5 is:
   \[
   1024 - 673 = \boxed{351}
   \]