Prove that infinitely many numbers in the sequence
{2012, 20122012, 201220122012, 2012201220122012,....} are multiples of 101.
By the way, there is nothing at all special about these values.
For any integer values of a, b and c (where c <> 0), the series s defined by
s(1) = a
s(n) = a + b*s(n-1)
includes infinitely many multiples of c.
Yes, any value of c! The series given above contains an infinite number of multiples of 678 and 123498765 and any integer (other than 0) that you might care to choose.
See my first post, which applies for all values of a, b and c. It has nothing to do with factoring a, b or c, or anything being relatively prime to anything else. As is often the case, proving the general result is easier than proving the specific result (because there are fewer distractions).
Generalizing a lot
