Erased digits

If we call the missing digits $x$ (2nd) and $y$ (15th) then we can calculate the alternating and the normal digit sum $s_a$ and $s_n$ .

$\begin{aligned} s_a &= 1-x+1-6+4-5+1-0+0-4+0-8+8-3+y-0+0-0 = -11-x+y \\ s_n &= 1+x+1+6+4+5+1+0+0+4+0+8+8+3+y+0+0+0 = 41+x+y \end{aligned}$

Since $19!$ is divisible by $9$ and $11$ , these digit sums have to be multiples of those numbers.

$-11-x+y$ can only be a multiple of 11 for $x=y$ , because $0 \leq x,y \leq 9$ .

$41+x+y = 41+2x$ can be at most $41+2\cdot 9 = 59 < 63$ . Note that it also has to be odd since we're adding odd (41) and even ( $2x$ ). The only possible solution is $41+2x=45 \Rightarrow \boxed{x=y=2}$

We note that the number of trialing zeros in $19!$ is $\left \lfloor \dfrac {19}5 \right \rfloor = 3$ . This means the last three digits (16th to 18th) are zeros. Then the 15th digit $d_{15}$ is given by:

$\begin{aligned} \frac {19!}{10^3} & \equiv d_{15} \text{ (mod 10)} \\ & \equiv 19 \times 18 \times 17 \times 16 \times {\color{#3D99F6}\cancel {15}^3} \times 14 \times 13 \times 12 \times 11 \times {\color{#3D99F6}\cancel {10}^2} \times 9 \times {\color{#D61F06}\cancel{8}^1} \times 7 \times 6 \times {\color{#3D99F6}\cancel {5}^1} \times 4 \times 3 \times 2 \text{ (mod 10)} & \small \color{#D61F06} \text{Since }2^3 = 8 \\ & \equiv 9 \times 8 \times 7 \times 6 \times 3 \times 4 \times 3 \times 2 \times 2 \times 9 \times 7 \times 6 \times 4 \times 3 \times 2 \text{ (mod 10)} \\ & \equiv 2 \text{ (mod 10)} \end{aligned}$

Now we have $19! = \overline{1d_21645100408832000}$ . Since $19!$ is divisible by 9, its sum of digits is also divisible by 9. We note that the sum of digit without $d_2$ is $1+1+6+4+5+1+4+8+8+3+2 = 43$ . The next multiple of 9 is 45 then $d_2 = 2$ . The one after is 54 then $d_2 = 11$ , which is unacceptable. Therefore, $d_2 = 2$ and $d_{15}=2$ and the answer is $\boxed{22}$ .

Since 19! ends with three zeroes, we can divide it by 1000 and still get an integer:

$\frac{19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{10^3} =$

$\frac{19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}{2^3 \times 5^3} =$

$19 \times 18 \times 17 \times 16 \times \frac{15}{5} \times 14 \times 13 \times 12 \times 11 \times \frac{10}{5} \times 9 \times \frac{8}{2^3} \times 7 \times 6 \times \frac{5}{5} \times 4 \times 3 \times 2 =$

$19 \times 18 \times 17 \times 16 \times 3 \times 14 \times 13 \times 12 \times 11 \times 2 \times 9 \times 7 \times 6 \times 4 \times 3 \times 2$

Finding this product's last digit is the same as finding the original number's 15th digit, so we'll evaluate the product mod 10:

$19 \times 18 \times 17 \times 16 \times 3 \times 14 \times 13 \times 12 \times 11 \times 2 \times 9 \times 7 \times 6 \times 4 \times 3 \times 2 \pmod{10} \equiv$

${\color{#D61F06}9 \times 8} \times {\color{#3D99F6}7 \times 6} \times {\color{#20A900} 3 \times 4 \times 3 \times 2} \times 1 \times {\color{#69047E} 2 \times 9} \times {\color{cyan} 7 \times 6} \times {\color{magenta} 4 \times 3 \times 2} \pmod{10} \equiv$

${\color{#D61F06}72} \times {\color{#3D99F6}42} \times {\color{#20A900} 72} \times {\color{#69047E} 18} \times {\color{cyan} 42} \times {\color{magenta} 24} \pmod{10} \equiv$

$2 \times 2 \times 2 \times 8 \times 2 \times 4 \pmod{10} \equiv$

$512 \pmod{10} \equiv$

$2 \pmod{10}$

Therefore, the 15th digit of 19! is 2.

Since in 19! is a multiple of nine, the sum of its digits must also be a multiple of nine, so if we call the 2nd digit $x$ :

$1+x+1+6+4+5+1+0+0+4+0+8+8+3+2+0+0+0 =$

$43 + x$

And the only digit that makes the sum divisible by nine is $2$

So the 2nd digit is $\boxed{2}$ and the 15th digit is $\boxed{2}$