Interesting problem on determinant (and pinch of number theory)

Question

The digits A, B and C are such that the three digit numbers A88 , 6B8 and 86C are divisible by 72, what is $$\begin{vmatrix} A & 6 & 8\\ 8 & B & 6 \\ 8 & 8 & C \end{vmatrix} \pmod {72}$$

I can (and did) find the individual digits B and C , A was a bit harder to find. And then solved the expanded the Determinant. This was quite time consuming.

Is it possible to solve this solely using properties of determinant (or matrix)? Edit: without actually finding the digits

Answer 1

Since replacing a row in a matrix by that same row plus a multiple of another row does not change the determinant, we see that by replacing the third row by the third row plus ten times the second row plus a hundred times the first row... that is... by performing the modification:

$$R_3 \leftarrow R_3 + 10R_2+100R_1$$

we arrive at a matrix who should have the same determinant as before. The new third row however has every entry divisible by $72$, as per our hypothesis, noting that the entries in our new third row are precisely those three-digit numbers $A88, 6B8$ and $86C$

As such, it follows that the determinant itself is divisible by $72$ and is then equivalent to $0$ when considered modulo $72$.

Answer 2

Once we have $A,B,C$ we are after $$\begin{vmatrix} 2 & 6 & 8\\ 8 & 4 & 6 \\ 8 & 8 & 4 \end{vmatrix} \pmod {72}$$ We note we can divide each row by $2$, which shows the determinant is a multiple of $8$. If we do the division we can work $\bmod 9$, so we now want $$\begin{vmatrix} 1 & 3 & 4\\ 4 & 2 & 3 \\ 4 & 4 & 2 \end{vmatrix} \pmod {9}$$ Now we can add the first row to the third and the second row to the third, getting a row of $0$s. The original determinant is then $0 \pmod {72}$