<s>[INST] 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.
 [/INST] 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 $1a0767436....$
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