Given the determinant of order $n$
$$\begin{vmatrix}
   8   &  3   &  3  &  \dots  & 3 \\
   3   &  8   &  3  &  \dots  & 3 \\
   3   &  3   &  8  &  \dots   & 3 \\
\vdots &  \vdots  &  \vdots &  \ddots & \vdots \\
   3   &  3   &  3  &  \dots  & 8
\end{vmatrix}$$
Calculate its value and determine for which values of $n$ this value is a multiple of $10$.
To solve the given problem, we need to compute the determinant of the matrix and determine for which values of \( n \) this determinant is a multiple of 10. The matrix is of order \( n \) and has the form:

\[
\begin{vmatrix}
8 & 3 & 3 & \dots & 3 \\
3 & 8 & 3 & \dots & 3 \\
3 & 3 & 8 & \dots & 3 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
3 & 3 & 3 & \dots & 8
\end{vmatrix}
\]

We will use the general formula for the determinant of a matrix where the diagonal elements are \( a \) and the off-diagonal elements are \( b \). For our specific case, \( a = 8 \) and \( b = 3 \).

1. **General Determinant Formula**:
   The determinant of such a matrix can be expressed as:
   \[
   \boxed{(a - b)^{n-1} (a + (n-1)b)}
   \]
   Substituting \( a = 8 \) and \( b = 3 \), we get:
   \[
   \boxed{(8 - 3)^{n-1} (8 + (n-1) \cdot 3)}
   \]

2. **Simplify the Expression**:
   \[
   (8 - 3)^{n-1} (8 + 3(n-1)) = 5^{n-1} (8 + 3n - 3) = 5^{n-1} (3n + 5)
   \]

3. **Determine When the Determinant is a Multiple of 10**:
   For the determinant to be a multiple of 10, the expression \( 5^{n-1} (3n + 5) \) must be divisible by 10. Since \( 5^{n-1} \) is always divisible by 5 for \( n \geq 1 \), we need \( 3n + 5 \) to be divisible by 2 (i.e., \( 3n + 5 \) must be even).

4. **Solve for \( n \)**:
   \[
   3n + 5 \equiv 0 \pmod{2}
   \]
   Simplifying, we get:
   \[
   3n \equiv -5 \pmod{2} \implies 3n \equiv 1 \pmod{2} \implies n \equiv 1 \pmod{2}
   \]
   Therefore, \( n \) must be odd.