This riddle bothers me for a few weeks now and I'm starting to worry that I need some $p$-adic Number theory to solve this. I solve most of the riddles in a day, but this one is just annoying to me. I was thinking of taking the number $10!$. Any help is appreciated!
Edit I'm looking for a number that is written as only a series of $1$'s and $0$'s which represent the number $\sum_{n=0}^{\infty}a_n*p^n$ with $a_n\in\{0,\ldots,p-1\}$ where $a_n$ is either $1$ or $0$.
Edit 2 Probably this is a not so well known problem, but people are working on it with no solution yet found for this case. They let a program search for such numbers and there were no matches so far. There is a solution in base $1,\;\ldots,\;5$, namely $$82000 = 10100000001010000\ (2) = 11011111001\ (3) = 110001100\ (4) = 10111000\ (5).$$ This number fails in the original problem for $82000=1431344\ (6)$.
Originally checked to $2^{65520}$ (or about $3*10^{19723}$) on Nov 07 2008.
Conjectured to be complete. a(4), if it exists, it is greater than $10^{15}$. Apr 06 2012.
In 2016, a Mathematician stated that it is plausible that there are no more such terms, but it is not proven.
See for yourself: https://oeis.org/A146025.
Some heuristic argument (too long for a comment):
Suppose we have an $n$-digit number $m$ in base 10 that can be represented by 0s and 1s. There are $2^{n-1}$ such $m$s. Written in a different basis $B$, that number has approximately $n / \log B$ digits where $\log$ denotes the logarithm to base 10.
Assuming that the digits of $m$ in other bases are independent, the probability that a digit in base $B$ happens to be in $\{0,1\}$ is $2/B$. Thus, the probability that the representation is again all 0s and 1s in base $B$ is$$\left(\frac2B\right)^{n/\log B}$$
As the condition shall hold for more than one basis, we arrive at a probability of $$\prod_{4\leqslant B \leqslant 9} \left(\frac2B\right)^{n/\log B}$$ that $m$ meets the condition in all bases. Notice that the condition is met trivially for $B=2$, and when the condition is met for $B=9$, then it's also met for $B=3$.
The $\def\mod{\operatorname{mod}}$ conditions for $B=8$ and $B=4$ are not independent, though, because they are linked by the binary representation: If we expand the octal digits pair-wise in binary, $m$ has the following representation in groups of 6 bits: $$\cdots00b_i\,00b_j\cdots$$ Thus matching $B=4$ and $B=8$ means $b_i=0$ where $i=3\mod 6$. Only $b_j$ can be chosen (where $j=0\mod 6$). Which means we can replace the two bases 4 and 8 by $B=2^6=64$. Hence a corrected probability for $m$ to match is $$\prod_{B\in\{5, 6, 7, 9, 64\}} \left(\frac2B\right)^{n/\log B}$$ and the expected number of finds amongst the $n$-decimal digit numbers is $$E_n=2^{n-1}\!\!\!\!\!\!\!\prod_{B\in\{5, 6, 7, 9, 64\}} \left(\frac2B\right)^{n/\log B} =\left( 0.5\!\!\!\!\!\!\prod_{B\in\{5, 6, 7, 9, 64\}} \left(\frac2B\right)^{1/\log B} \right)^n =:a^n $$ This gives a value of $$ a \approx 0.000227 $$ with the following contributions of the $B$s:
Under the assumption that the $E_n$s are independent, the expected value of solutions with $n_0$ digits or more is $$ \sum_ {n\geqslant n_0}E_n =\sum_{n\geqslant n_0}a^n = a^{n_0} \sum_{n=0}^\infty a^n = a^{n_0} \frac1{1-a} \approx a^{n_0} $$ where the approximation assumes $|a|\ll 1$ which hols for our $a$.
According to OEIS.A146025, there is no solution for our $B$s below $3\cdot 10^{19723}$ (except the trivial ones 0 and 1), and therefore we get an estimate using $n_0=19723$: $$ E_{\geqslant 19723} \approx 0.000227^{19723}\approx 10^{-71870} $$
Conclusion: If I had to bet whether there exists a solution to your problem, I wouldn't even bet a flower pot — except I wanted to get rid of that pot real quick.