Prove that infinitely many numbers in the sequence
{2012, 20122012, 201220122012, 2012201220122012,....} are multiples of 101.
Consider the sequence, modified by adding a 0 as the first term:
{0, 2012, 20122012, 201220122012, 2012201220122012,....}
Note that Term n+1 = Term n*10000 + 2012
Consider the sequence a(n) = mod(term n, 101).
a(1) = mod(0,101) = 0
a(n+1) = mod( 10000*a(n) + 2012, 101).
This is necessarily a repeating sequence with some cycle, and because it includes 0, it must necessarily include an infinite number of zeroes. Therefore, the sequence with 0 added has an infinite number of terms which are multiples of 101, and thus the original sequence has also.
This concludes the proof.
However, I did use Excel to discover that Term 101 and Term 202 are both divisible by 101. And we already know about term 0. The length of the repeating cycle is 101, so there are an infinite number of terms in the sequence with every residue mod 101.
Edited on April 30, 2012, 9:32 pm
No problem (spoiler)
