Ytukyg

$$15! equiv 1cdot 2cdot 3cdot,cdots,cdot 15 equiv 1square0767436square000$$

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:
$$15! approx sqrt{2picdot 15}
cdot left(frac{15}{e}right)^{15}$$
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.

Another way to reason is to note that $15!$ is divisible by $2cdot4cdot2cdot8cdot2cdot4cdot2=2^{11}$, which means $1square0767436square$ is divisible by $2^8$. In particular, it's divisible by $8$. But since $8mid1000$ and $8mid360$, 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
$$
15!=2^{11}cdot 3^6cdot 5^3cdot 7^2cdot11cdot13=1000cdot 2^8cdot3^6cdot7^2cdot 11cdot 13
$$
this means finding the remainder modulo $10$ of
$$
2^8cdot3^6cdot7^2cdot 11cdot 13
$$
that gives $8$ with a short computation.

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

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

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

$1*2*3*4*6*7*9*2*11*12*13*14*3equiv 2*3*4*(-4)*(-3)*(-1)*2*1*2*3*4*3equiv$

$-2^9*3^4 equiv -512*81equiv -2 equiv 8pmod {10}$..

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 $0le a le 9$ 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 $1a0767436....$

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 $d_1=3$.

$15!=2^{11}cdot 5^3cdot 7^2cdot 11cdot 13=(1000)X$ where $X=2^8cdot 3^6cdot 7^2cdot 11cdot 13.$

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 $6cdot 9cdot 9 cdot 1cdot 3equiv 6cdot(-1)^2cdot 3equiv 18equiv 8$. 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 $9$.