2001 OIM Problems/Problem 1
Problem
We say that a natural number 
is "çharrúa" if it simultaneously satisfies the following conditions:
- All digits of
are greater than 1.
- Whenever four digits of are multiplied, a divisor of is obtained.
Show that for each natural number 
there is a çharrúa number with more than digits.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
Clearly if we show that there are infinite such numbers, then the criterion is satisfied. Thus we create such a number. Define
![\[X=22223232\]](http://latex.artofproblemsolving.com/f/6/7/f6731ce23b645584870f6d3ad1f298062b707840.png)
Notice that 
clearly has 
as a factor and also has 
as a factor by the sum of the digits; therefore, this number is çharrúa.
Next, let 
be some natural number divisible by 
, and add 
's to the beginning of as digits. No additional possibilities for the product of four digits are introduced, and the resulting number is divisible by by construction as well as 
from before, and since can be any positive multiple of , the conclusion follows.
