<s>[INST] 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.
 [/INST] 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 $x=4$.
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