Prove that in the Fibonacci sequence (0, 1, 1, 2, 3, 5, 8, 13, ... where each number is the sum of the two previous) there's at least one number that ends in 999999.
Let a0 = 0, a1 = 1, and so on. Notice that a-1 = 1 and a-2 = -1.
Consider the sequence, mod 106.
The number of distinct consecutive pairs of terms is 1012, and so at some point we will get a duplicate pair. (Pigeonhole principle.)
Let's say ak = ak+n and ak+1 = ak+1+n.
It follows that ak-1 = ak+n-1, ak-2 = ak+n-2, ... a0 = an, and a-2 = an-2.
Hence an-2 will end in 999999.