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.

Let $d_1$ and $d_2$ be the two unknown digits.
The number must be divisible by $8000$, because $15!$ contains $8$ and $1000$.
$d_2$ is non-zero number, because $15!$ contains only three $5$s. It implies $1d_10767436d_2$ must be divisible by $8$. It implies $36d_2$ is divisible by $8$. Hence, $d_2=8$.
Now you can use the divisibility by $9$ ($d_1+d_2=11$) and find $d_1=3$.
