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 a_{0} = 0, a_{1} = 1, and so on. Notice that a_{-1} = 1 and a_{-2} = -1.

Consider the sequence, mod 10^{6}.

The number of distinct consecutive pairs of terms is 10^{12}, and so at some point we will get a duplicate pair. (Pigeonhole principle.)

Let's say a_{k} = a_{k+n} and a_{k+1} = a_{k+1+n}.

It follows that a_{k-1} = a_{k+n-1}, a_{k-2} = a_{k+n-2}, ... a_{0} = a_{n}, and a_{-2} = a_{n-2}.

Hence a_{n-2} will end in 999999.