Show that there are infinitely many integers n such that:

1) All digits of n in base 10 are strictly greater than 1.

2) If you take the product of any 4 digits of n, then it divides n.

(10^27-1)/3 is a number which fits the conditions of the problem. (10^27-1)/3 = 333333333333333333333333333 which is a multiple of 3*3*3*3.

An infinite sequence can be made using (10^(27*n)-1)/3