Generate a $5 × 5$ matrix such that the each entry is an integer between $1$ and $9$, inclusive, and whose determinant is divisible by $271$. 
Generate a $5 × 5$ matrix such that the each entry is an integer between $1$ and $9$, inclusive, and whose determinant is divisible by $271$.

This is a practice problem for a linear algebra exam I have coming up, and can't for the life of me figure it out. I was thinking maybe making a triangular matrix ($0$s below main diagonal, determinant would be the product of the diagonal), but that wouldn't work because it says to use integers $1$-$9$.
Been thinking of this problem all day. The only way we've covered the determinant for a large matrix ($3 × 3$ or larger) has been through summing up the signed elementary product, but for a $5 × 5$ matrix, you'd need to make sure that all $5! = 120$ signed elementary products would need to be divisible by $271$.
If anyone has a better way to approach and solve this problem, it would be very much appreciated.
 A: 
If you make all the entries $1$, then the determinant is $0$, which is divisible by $271$. 

courtesy of @ Daniel Schepler
A: Let $\overline{a_{i1}a_{i2}a_{i3}a_{i4}a_{i5}}$, $i \leq 1 \leq 5$ be five distinct five digit multiples of $271$ which have non-zero digits. We claim that the matrix $a_{ij}$ has a determinant divisible by $271$. That is, if we have the matrix whose rows are the multiples of $271$, then it has a determinant divisible by $271$.
The proof is rather clever, really : note that determinants are preserved under column addition and subtraction operations. Therefore, if $C_i$ are the columns, perform $C_5 \to 10000C_1 + 1000C_2 + 100C_3 + 10C_4 + C_5$. Performing this, the determinant does not change, but now, $(C_5)_j = 10^4 a_{j1} + 10^3 a_{j2} + .. = \overline{a_{j1} a_{j2}...a_{j5}}$, for all $1 \leq j \leq 5$. Hence, the column $C_5$ consists completely of multiples of $271$. By multilinearity of the determinant, it follows that the determinant of the matrix is divisible by $271$.
I am sure that you can easily find five multiples of $271$ which satisfy the situation, and furthermore give a non-zero determinant, divisible by $271$.
Take these multiples : $99728,98915,66666,31436,48238$, and form the matrix:
$$
\begin{pmatrix}
9\quad 9 \quad 7 \quad 2 \quad 8 \\
9 \quad 8 \quad 9 \quad 1 \quad 5 \\
6 \quad 6 \quad 6 \quad 6 \quad 6 \\
3 \quad 1 \quad 4 \quad 3 \quad 6 \\
4 \quad 8 \quad 2 \quad  3\quad 8\\
\end{pmatrix}
$$
which has determinant $-1626 = 271 \times -6$. Interestingly enough, if you write $11111$ above instead of $66666$, and swap two rows, you will get exactly $271$ as the determinant.
