Determining the missing digits of $15! \equiv 1\square0767436\square000$ without actually calculating the factorial

Another way to reason is to note that $15!$ is divisible by $2\cdot4\cdot2\cdot8\cdot2\cdot4\cdot2=2^{11}$, which means $1\square0767436\square$ is divisible by $2^8$. In particular, it's divisible by $8$. But since $8\mid1000$ and $8\mid360$, the final $\square$ must be either $0$ or $8$. But it can't be $0$, since $15!$ has only three powers of $5$ (from $5$, $10$, and $15$), and those were already accounted for in the final three $0$'s of the number. Thus the final $\square$ is an $8$. Casting out $9$'s now reveals that the first $\square$ is a $3$.

Remark: It wasn't strictly necessary to determine the exact power of $2$ (namely $2^{11}$) that divides $15!$, merely that $2^6$ divides it, but it wasn't that hard to do.

You can cast out $9$’s and $11$’s: \begin{align} 1+x+0+7+6+7+4+3+6+y+0+0+0&=x+y+34 \\ 1-x+0-7+6-7+4-3+6-y+0-0+0&=-x-y \end{align} Thus $x+y=11$ (it can't be $x=y=0$).

Then find the remainder modulo $10000$; since $$ 15!=2^{11}\cdot 3^6\cdot 5^3\cdot 7^2\cdot11\cdot13=1000\cdot 2^8\cdot3^6\cdot7^2\cdot 11\cdot 13 $$ this means finding the remainder modulo $10$ of $$ 2^8\cdot3^6\cdot7^2\cdot 11\cdot 13 $$ that gives $8$ with a short computation.

Using the divisibility rule for 7 the answer boils down to 3 and 8:

$-368+674+307+1 \mod 7 \equiv 0$