Find $x$ and $y$ where $20!=\overline{24329020081766xy\dots}$ Find $x$ and $y$ where $20!=\overline{24329020081766xy\dots}$(without using calculator.)
My attempt:I first find how many zeroes does it have:
$$\left\lfloor {\frac{20}{5}} \right\rfloor=4.$$
It can be solved easily if we know that after $y$ there are only three digits then we can know: 
$$y=0.$$
Then $\overline {6x}$ is divisible by $4$ which gives us:
$$x=4\ \ \ \text{ or }x=8.$$
Then if we check divisiblity role of $8$ we will get that $\overline {66x}$ is divisible by $8$ that tells to us $x$ can only be $4$. Thus
$$x=4.$$
But know the biggest problem is that we don't know how many digits are there after $y$.
Or in a bigger amount how many digits are there in $20!$.
Thanks.
 A: 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 $x=4$.
A: 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.
A: The number of digits can be found using logarithm. You can use a logarithm table to find the number of digits without calculator:
$$\log_{10}(20!)=\sum\limits_{i=1}^{20}\log_{10}(i)\approx 18.38 $$
Therefore that number has 19 digits!
A: You are almost there.  It is also true that $20!$ is divisible by $9$, so the sum of the digits is divisible by $9$ as well.  That gives you $x$
