Determining the missing digits of $15! \equiv 1\square0767436\square000$ without actually calculating the factorial 
$$15! \equiv 1\cdot 2\cdot 3\cdot\,\cdots\,\cdot 15 \equiv 1\square0767436\square000$$
Using a calculator, I know that the answer is $3$ and $8$, but I know that the answer can be calculated by hand.

How to calculate the missing digits? I know that large factorials can be estimated using Stirling's approximation:
$$15! \approx \sqrt{2\pi\cdot 15}
\cdot \left(\frac{15}{e}\right)^{15}$$
which is not feasible to calculate by hand.
The resulting number must be divisible by 9 which means that the digit sum must add up to 9 and is also divisible by 11 which means that the alternating digit sum must be divisible by 11:
$1+ d_0 + 0 + 7 +6 +7 +4 +3+6+d_1+0+0+0 \mod \phantom{1}9 \equiv \,34 + d_0 + d_1 \mod \phantom{1}9 \equiv 0 $
$-1+ d_0 - 0 + 7 -6 +7 -4 +3-6+d_1-0+0-0 \mod 11 \equiv d_0 + d_1 \mod 11 \equiv 0 $
The digits $3$ and $8$, or $7$ and $4$, fulfill both of the requirements.

$15!=2^{11}\cdot 5^3\cdot 7^2\cdot 11\cdot 13=(1000)X$ where $X=2^8\cdot 3^6\cdot 7^2\cdot 11\cdot 13.$
The last digits of $2^8(=16^2), 3^6 (=9^3),7^2, 11,13$ are, respectively $6,9,9,1,3 .$
Modulo $10$ we have $6\cdot 9\cdot 9 \cdot 1\cdot 3\equiv 6\cdot(-1)^2\cdot 3\equiv 18\equiv 8$. So the last digit of $X$ is an $8$. 
Therefore the  2nd digit of $15!$ must be a $3$ in order for the sum of all its digits to be divisible by $9$.
