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 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, ... a₀ = a_n, and a_-2 = a_n-2.

Hence a_n-2 will end in 999999.

Posted by Nick Hobson on 2004-06-15 15:36:20