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.
 A: 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.
A: 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.
A: Using the divisibility rule for 7 the answer boils down to 3 and 8:
$-368+674+307+1 \mod 7 \equiv 0$
A: Okay, $15! = 1*2*3..... *15=1a0767436b000$.
Why does it end with $000$?  Well, obviously because $5,10,15$ all divide into it so $5^3$ divides into it and at least three copies of $2$ divid into it so $2^3*5^3 =1000$ divide into it.
If we divide $15!$ by $100 = 8*5^3$ we get 
$1a0767436b = 1*2*3*4*6*7*9*2*11*12*13*14*3$
If we want to find the last digit of that we can do
$1a0767436b \equiv b \pmod {10}$ and 
$1*2*3*4*6*7*9*2*11*12*13*14*3\equiv 2*3*4*(-4)*(-3)*(-1)*2*1*2*3*4*3\equiv$
$-2^9*3^4 \equiv -512*81\equiv -2 \equiv 8\pmod {10}$..
So $b = 8$.
But what is $a$?
Well, $11|1a0767436b$ and $9|1a0767436b$.
So $1+0+6+4+6 - a - 7-7-3-b = 11k$ for some integer $k$. And $1+a+0+7+6+7+4+3+6+b = 9j$ for some integer $j$.
So $-a -8 =11k$ so as $0\le a \le 9$ we have $a = 3$.
And that's that $15! = 1307674368000$..... IF we assume the person who asked this question was telling the truth.
We do know that $15!$ ends with $.... 8000$ but we are completely taking someone elses word for it that it begins with $1a0767436....$
A: 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 $d_1=3$.
A: $15!=2^{11}\cdot 5^3\cdot 7^2\cdot 11\cdot 13=(1000)X$ 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 $9$.
