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≥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 5mod10, there is a difference of five terms. Following from this we see that the third appearance of 5mod10 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? vectorlinear algebramatrixPutnammodular arithmeticleast common multiplenumber theory