Find $x$ and $y$ where $20!=\overline{24329020081766xy\dots}$

Note that \begin{align}20! & = 2^{18}\cdot 3^8 \cdot 5^4 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \\ & = 2^{18} \cdot 3^8 \cdot 5^4 \cdot 7^2 \cdot (15-4) \cdot (15+4) \cdot (15-2) \cdot (15+2) \\ & = 2^{18} \cdot 3^8 \cdot 5^4 \cdot 7^2 \cdot (15^2 - 4^2) \cdot (15^2 - 2^2) \\ & < 2^{18} \cdot 3^8 \cdot 5^4 \cdot 7^2 \cdot 15^2 \cdot 15^2 \\ & = 2^{18} \cdot 3^{12} \cdot 5^8 \cdot 7^2 \\ & = 2^9 \cdot 9^6 \cdot (2 \cdot 5)^8 \cdot (2 \cdot 7^2) \\ & < 1000 \cdot 10^6 \cdot 10^8 \cdot 100 \\ & = 10^{19}.\end{align}

Therefore $20!$ has at most 19 digits.

Since $20!$ is divisible by $10000$, last four digits are zero.

We are given first $14$ digits of $20!$ (the last digit given being nonzero) and we know that last four digits are zero. Therefore $20!$ has $18$ or $19$ digits. At this point we see that $y=0$.

If $20!$ had 18 digits, then it would be equal to $243290200817660000$. However, this number is not divisible by $9$ (you can verify this by checking the sum of digits). Therefore $20!$ has exactly $19$ digits. Now, using the sum-of-digits-must-be-divisible-by-9 test again we conclude that $x=4$.

Some judicious pairing of the numbers from $1$ to $20$ leads to the rough estimate

$$\begin{align} 20!&=(20\cdot1)(17\cdot3)(19\cdot2)(12\cdot10)(15\cdot4)(18\cdot5)(16\cdot6)(14\cdot7)(13\cdot8)(11\cdot9)\\ &\approx20\cdot50\cdot40\cdot120\cdot60\cdot100\cdot100\cdot100\cdot100\cdot100\\ &\approx1000\cdot5000\cdot60\cdot10^{10}\\ &=10^3\cdot300000\cdot10^{10}\\ &=3\cdot10^{18} \end{align}$$

Some extra work could probably establish rigorous upper and lower bounds

$$2\cdot10^{18}\lt20!\lt3\cdot10^{18}$$

Knowing that $20!$ ends with $4$ $0$'s and counting that there are $14$ digits before the $xy$, we see that $y$ is the first of the trailing $0$'s. Finally, knowing that $9$ divides $20!$, we have

$$(2+4+3)+2+(9+0)+2+(0+0+8+1)+7+6+6+x\equiv5+x\equiv0\mod9$$

implies $x=4$.

Added later: Here is a second approach, using that fact that $20!$ is divisible by both $9$ and $11$.

We have

$$20!\lt20^{10}\cdot10!\lt2^{10}\cdot10^{10}\cdot(4\cdot10^6)=4096\cdot10^{16}\lt5\cdot10^{19}$$

so $20!$ has at most $20$ digits. Since it ends in $4$ $0$'s, everything to the right of the $xy$ is a $0$. Using the digit sum test for divisibility by $9$, we get

$$x\equiv4-y\mod9$$

Using the alternating digit sum test for divisibility by $11$, we get

$$x\equiv4+y\mod 11$$

The restriction $0\le x,y\le9$ makes it easy to check that $x=4, y=0$ is the only solution.