$10^{\phi (9^4)n}-1$ is charrua for every $n$.

carlosbr wrote: We say that a natural number $n$ is charrua if it satisfy simultaneously the next conditions: - Every digit of $n$ is greater than 1. - Every time that four digits of $n$ are multiplied, it is obtained a divisor of $n$ Show that every natural number $k$ there exists a charrua number with more than $k$ digits. Lemma: For all $n\in \mathbb {N}$ there is a $N\in \mathbb {N}$ such that $2^n|N$, $10^{n-1}\leq N<10^n$ and if $x$ is a digit on $N$, then $x\in \{1,2\}$

We shall prove that $2a_{k+3}$ is a charrúa number for $k$. In fact: Every digit of $2a_{k+3}$ belongs to $\{2,4\}$. The product of four digits of $n$ can be equal to $2^4,2^5,2^6,2^7,2^8$. Because $k\ge 4$ it holds that $2^8|2^{k+4}|2a_{k+3}$. $2a_{k+3}$ has $k+3>k$ digits. And we finish the proof. $\blacksquare$

Take the number $M=2222....223232$ where the amount of two's is of the form $9n+6$ and there's only 2 three's. Denote by $S(n)$ the sum of the digits of $n$. It is well knows that $9$ divides $n$ iff $9$ divides $S(n)$. Note that $S(M)= 2(9n+6)+3+3 = 18n+18 =>$ $9$ divides $M$. It is also well knows that $2^k$ divides $n$ iff $2^k$ divides the last $k$ digits of $n$. So $2^4$ divides $M$ iff $2^4$ divides $3232$ which is true since $3232= 202*16$ Therefore $3^2*2^4$ divides $M$. Now the product of any $4$ digits of $M$ is of the form $2^a*3^b$ with $4\ge a \ge 2$ and $2 \ge b\ge0$. So the product of any four digits of $M$ divides $3^2*2^4$ and therefore, the product of any four digits of $M$ divides $M$. Therefore $M$ is charrua. For any positive integer $k$ take $n=k$ and $M$ will have $9k+8$ numbers.

By the way, see here http://en.wikipedia.org/wiki/Charr%C3%BAa what charrúa means - the contest chairperson was maybe a descendant of them ... By the way, any number with prime divisors different from $2$ and $5$ has infinitely many repunit multiples (numbers made of just digit $1$). The proof is folklore. Thus $3^4$, $7^4$ and $9^4$ do that, and then those repunits multiplied by $3$, $7$, respectively $9$, are charrúa. So we can do this for any other number than four digits. The other solutions here are welcome if we impose as extra condition that not all digits be equal in a charrúa number.

it suffices to consider $22...23232$(9k+6 2s in total).

Motivated by the divisibility test of 9, by Euler's theorem, $n=\phi (243)=162$ satisfies $10^n \equiv 1(mod 243)$ so if we consider the number $333..333$ with 162 digits we can rewrite it as $3\cdot \frac{10^{162}-1}{9}$ meaning this number is divisible by $3^4=27$. So it is sufficient to pick a number of this form such that the number of its digits is a multiple of $162$.

$2222\dots2672$ with $6k$ $2$'s before the 6.