Deﬁne the determinant $D_1$ = $|1|$, the determinant $D_2$ =     
$|1 1|$ 
$|1 3|$      
, and the determinant $D_3=$        
|1 1 1|
|1 3 3|
|1 3 5|         
.
In general, for positive integer n, let the determinant $D_n$ have 1s in every position of its ﬁrst row and ﬁrst
column, 3s in the remaining positions of the second row and second column, 5s in the remaining positions of the third row and third column, and so forth. Find the least n so that $D_n$ $\geq$ 2015. 
1. We start by defining the determinant \( D_n \) as described in the problem. For example, \( D_1 = |1| \), \( D_2 = \begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix} \), and \( D_3 = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 3 & 3 \\ 1 & 3 & 5 \end{vmatrix} \).

2. To simplify the calculation of \( D_n \), we perform row operations. Specifically, we subtract the first row from all subsequent rows. This operation does not change the determinant of the matrix.

3. After subtracting the first row from all other rows, the matrix \( D_n \) transforms into an upper triangular matrix. The first row remains as \( [1, 1, 1, \ldots, 1] \), and the remaining rows will have the form \( [0, 2, 2, \ldots, 2] \), \( [0, 0, 2, \ldots, 2] \), and so on.

4. Next, we subtract the second row from all rows below it, then the third row from all rows below it, and so forth. This results in a matrix where the diagonal entries are all 2s, and all entries above the diagonal are also 2s.

5. The resulting matrix is an upper triangular matrix with 1s in the first row and 2s on and above the diagonal in all other rows. The determinant of an upper triangular matrix is the product of its diagonal entries.

6. Therefore, the determinant \( D_n \) is given by:
   \[
   D_n = 2^{n-1}
   \]

7. We need to find the smallest \( n \) such that \( D_n \geq 2015 \). Thus, we solve the inequality:
   \[
   2^{n-1} \geq 2015
   \]

8. Taking the logarithm (base 2) of both sides, we get:
   \[
   n-1 \geq \log_2(2015)
   \]

9. Using a calculator, we find:
   \[
   \log_2(2015) \approx 10.97
   \]

10. Therefore:
    \[
    n - 1 \geq 10.97 \implies n \geq 11.97
    \]

11. Since \( n \) must be an integer, we round up to the nearest whole number:
    \[
    n \geq 12
    \]

Conclusion:
The least \( n \) such that \( D_n \geq 2015 \) is \( \boxed{12} \).