MathDB

Hey, This problem is from the VTRMC 2006. 3. Recall that the Fibonacci numbers

F(n)

are defined by F(0) \equal{} 0, F(1) \equal{} 1 and F(n) \equal{} F(n \minus{} 1) \plus{} F(n \minus{} 2) for

n \geq 2

. Determine the last digit of

F(2006)

(e.g. the last digit of 2006 is 6). As, I and a friend were working on this we noticed an interesting relationship when writing the Fibonacci numbers in "mod" notation. Consider the following, 01 = 1 mod 10 01 = 1 mod 10 02 = 2 mod 10 03 = 3 mod 10 05 = 5 mod 10 08 = 6 mod 10 13 = 3 mod 10 21 = 1 mod 10 34 = 4 mod 10 55 = 5 mod 10 89 = 9 mod 10 Now, consider that between the first appearance and second apperance of

5 mod 10

, there is a difference of five terms. Following from this we see that the third appearance of

5 mod 10

occurs at a difference 10 terms from the second appearance. Following this pattern we can create the following relationships. F(55) \equal{} F(05) \plus{} 5({2}^{2}) This is pretty much as far as we got, any ideas?

Problem Statement